When you use the zero product property, it is usually wise to check the answer in the original equation. I would have tacked that on to the end of the video.
@@farhanrejwanIt’s absolutely not necessary since he used equivalences on each step, meaning he can also go back. Or in other words, he also proved that if x=the solution, then x satisfies the equation, at the same time while proving that if x is a solution then x=that solution he found
@@youssefchihab1613 i'm not entirely convinced. i've done such steps before, and the answers didn't satisfy the original question (that one probably had squares and roots, come to think of it).
Step 1: Find that 1 is a solution. Step 2: Imagine a graph of an exponential function to decide whether there is another solution. Step 3: Watch the video, because you can't solve the equation.
My inelegant solution involved converting 3 and 6 to 2^(ln3/ln2) and 2^(ln6/ln2) respectively, then an evil quadratic appears, and with a calculator get the answers 1 and -1.631..... Presh's solution is much nicer!
Got both solutions. Working through it using the quadratic formula, you get to a point where you're questioning reality: "I know that all of this square-root mess needs to simplify to 1 for one of the solutions, but how to get there...?" The key insight for my method was to notice that the property ln(6) = ln(2x3) = ln(2) + ln(3) will lead to a perfect square that will negate the square-root operation.
I did the same thing, except that I (somehow) failed to notice the x=1 solution before I started. But after stumbling across the same trick to simplify the sqrt term, I got (-log3(2) ± (log3(2) + 2)) / 2, and said "oh yeah duh" when one of the solutions was x=1. And then double-checked that the other solution actually worked.
Same here ... but it gave a monstrous delta and convoluted x1 and x2 ... didn't seem to me that I could simplify them to get Presh's solituons, was expecting at least one of them to easily turn to 1 !!
But you for forgot to consider the negative values. Despite x being negative x^2 is always positive. So in negatives while 2^x decreases, 3^x^2 inxreases, making your argument invalid for the general case. If the question only asks about positive real values, then your solution is a perfectly acceptable solution.
I knew there would've been an another solution besides x = 1 because of the x^2, but I knew it wouldn't be an integer or anything and I didn't wanna bother. Good video.
See, I just took the natural log of both sides at the start and got xln(2)+x^2 ln(3)=ln(6) converted that to x^2 ln(3)+xln(2)-ln(6)=0 and then used the quadratic formula. I can see your approach being the answer used on a math test where no calculator is available, but that's the route my mind immediately went to.
In all honesty ... yes your solution worked, but you really made a mess of it. For gods sake ... just take the log on both sides immediately ... ln(2^x * 3^x^2) = ln6 ln(2^x) + ln(3^x^2) = ln6 x ln2 + x^2 ln3 = ln6 or x^2 ln3 + x ln2 - ln6 = 0 use quadratic formula and you're done. Your solution gave you a way to factor it without quadratic formula but that is coincidence ... just replace the 6 on the right side by 7 and your solution is toast ... while the one shown here still works.
Or combine the two. Use your method, but also use the fact that at first glance you see that x=1 is a solution. That often helps solve quadratic formulas quicker. Then you see ln6 is ln3+ln2, you can get the factor (x-1) out and get the other factor in your head without using paper. Filling in the quadratic takes some time.
In high school, i had a teacher that would have problems like this on tests. Problems that were almost logically too easy to solve, but would then only award half a point if you didn't show your work.
There is a trick with the quadratic, x^2 log3 + x log2 - (log2 + log3) = 0. Since you know one solution is x=1 then the other is always x=c/a= - (log2 + log3)/log3 = -(1 + log2/log3). No need to do the full solution. This comes from c/a = root1*root2 for all quadratics.
Yeah weirdly they decided that a^b^2 = a^(b^2) instead of (a^b)^2 in this problem even though that goes against the standard convention. So I was like oh I see it's just one (although there could be another solution which there is here), but then I dismissed that remembering that a^b^2 is actually (a^b)^2 which make 1 an incorrect answer. Instead you would get to say a^b^2 = a^2b which would completely different than the problem he solved in the video. Suffice to say I find the conventions around exponents and when which ones get applied very confusing. If anyone can explain how you are supposed to see this a^b^2 and know it is a^(b^2) different than other times it is written like this but it would be a^b^2 is treated as (a^b)^2, please let me know.
@brcoutme stop trying to intuit some convention that doesn't exist and watch the video for proper context clues... he explains in words what the equation is actually supposed to be interpreted as. Without parentheses it is just illdefined.
Best method IMO is: spot that x = 1 works. Then use logs from the equation (not the long way like Presh) and get x^2 ln 3 + x ln 2 = ln 6, subtracting ln 6 from both sides to have 0 on the right. Then the knowledge that x - 1 must be a factor helps us factorise it as (x - 1)(x ln 3 + ln 6) = 0.
Honestly this one was pretty easy, also I am pretty sure there are other ways for solving for both solutions but yours was quite straightforward and simple👍🏼
At 3:20 you can also see that we have (2*3^(x+1))^(x-1)=0 Now we see that either x=1(trivial solution), or 2*3^(x+1) is 1(as it is positive) Then you take log and voila
Solving normally , the discriminant ((log2)^2+4(log3)(log6)) can be factorized to ((log2+2log3)^2). But its way easier to try and rearrange to factor out x-1 as we know one solution. The easiest solution according to me is using that x1+x2 = (-b/a) or x1 * x2 = (c/a) . since x1 = 1 its trivial to obtain x2
After figuring out that x=1 is a solution, we can do the following: 2^x * 3^(x^2) = 6 2^(x-1) * 3^(x^2-1) = 1 x^2-1 is a difference of squares 2^(x-1) * 3^((x+1) * (x-1)) = 1 Since x=1 was already found intuitively, you can raise to the power of 1/(x-1) without worrying about missing a solution 2 * 3^(x+1) = 1 2 * 3 * 3^x = 1 3^x = 1/6 x = log_3(1/6) x = ln(1/6)/ln(3) x = -ln(6)/ln(3)
Trying to substitute this back in using a phone calculator or various internet calculators is challenging! I think they can't cope with the second exponent.
2=3^a Ln 2 = a ln 3 a= ln 2 / ln3 (3^ax) × 3^x² = 6 3^(ax+x²)=3^b 6=3^b b=ln6/ln3 x²+ax=b x²+ax-b=0 Solve as a quadratic using the formula. Imma use a calc so its easier. x= 1 first solution Second solution is -ln6 /ln3. Ezpz
You can just log both sides straight away and use log rules to make a quadratic ((log(3)x^2 + (log(2)x - log(6) = 0), then use the quadratic equation to solve it for both values (1 and -1.6309…). That seems much easier in my opinion.
I solved this by expanding everything out and using log. Taking the log of both sides got me x^2 log(3) + x log(2) - log(6) = 0. Since we know 1 is a solution, we can factor (x-1) out. This gets us (x-1)(x log(3) + log(6)) = 0, and so x = 1 or x = -log(6)/log(3). Which are the solutions.
I stumbled upon a nice solution. We note that 1 is a solution. We also note that the equation can be reduced to a quadratic with leading and linear coefficients log(3) and log(2), respectively. Thus, since one of the solutions is 1, the other must be -1-(log(2)/log(3)), by Vieta's formulae. PS: log denotes the natural logarithmic function throughout but other bases would work equally well.
I used a modified form of the quadratic equation but couldn't simplify it easily. I didn't want to do it again, so I pushed through, called the unknown square root z, and finally simplified it by using the solution x=1. I was super proud of this creative route until I read that I could've just used polynomial division and didn't need any quadratic formula at all, which I gracefully forgot about. Sometimes, you can't see the big picture when you're so focused on the details
Very nice! I did the quadratic equation, then quickly used Horner's scheme to divide (x-1) out, and found the other linear factor to be (ln3 • x + ln6), so x = -ln6/ln3 = -1 - ln2/ln3. 🤓
I did ln(2^x * 3^x^2)=ln(2^x) + ln(3^x^2)=ln(3)x^2 + ln(2)x Then you do a quick rearrange to get ln(3) as a, ln(2) as b, and -ln(6) as c in the quadratic equation. Then you use wolfram alpha to tell you that the big messy +- quadratic equation is equal to 1 and -1.63
Can't you take the log of both sides? log(2^x * 3^(x^2)) = log(2^x) + log(3^{x^2)) = xlog2 + x^2log3 = log6 Let b = log2, a = log3, c = log6 and this becomes bx + ax^2 = c and then it's just a quadratic equation that can be solved using the quadratic formula? Am I missing something?
Yes, you can do that. That's how I did it. However, you'll see that you'll get a nasty determinant: √[(ln2)² + 4 ln3 ln6] It wasn't immediately obvious to me how to proceed her but I figured it out after a while. You'll have to rewrite ln6 as ln2 + ln3, and then proceed as follows (I'll only look at the argument under the root for simplicity's sake): (ln2)² + 4 ln3 ln6 = (ln2)² + 4 ln3 (ln2 + ln3) = (ln2)² + 4 ln2 ⋅ ln3 + 4 (ln3)² = (ln2)² + 2 ln2 ⋅ 2 ln3 + (2 ln3)² Now, we have a perfect square of ln2 and 2 ln3. 😀 (ln2)² + 2 ln2 ⋅ 2 ln3 + (2 ln3)² = (ln2 + 2 ln3)² Applying the square root obviously only leaves us with ln2 + 2ln3. After that, the two solutions of the quadratic formula collapse to either 1 or -(ln2 + ln3) / ln3 which can also be written as -ln6 / ln3. Knowing that things get a little tricky with the quadratic formula, Presh's solution turns out to be the easier approach, in my opinion, but I must admit that it hadn't occurred to me to do the steps he did in the beginning (i.e. dividing both sides by 6, splitting 6 into 2⋅3 and then work these factors into the exponents of the numerator).
Better still, if you choose the correct base (when you can choose the value of one coefficient, which value do you choose of which coefficient?) then it factorises straight away. x^2 + (log 2)x - log 6 = 0 x^2 + (log 6 - 1)x - log 6 = 0 (x + log 6)(x - 1) = 0 x = -log 6, x = 1
if you have a calculator you can just take the natural log of both sides, use properties of logs to split 2^x and 3^x^2 into addition, use more properties of ln to get the exponents to the front of the natural logs, subtract over and use the quadratic equation to get the solution immediately without all these hoops of logic.
I’d have just written both of the bases as e^ln(base), added the exponents together, taken the ln of both sides, and gotten a quadratic that you can probably just quad formula.
This was a good one. I used sympy to try to solve this equation but it threw up it's hands, giving this error: NotImplementedError: multiple generators [2**x, 3**(x**2)] No algorithms are implemented to solve equation 2**x*3**(x**2) - 6 I then used Wolfram Alpha and it gave x = {1, - log(6) / log(3) }. It is not immediately obvious how this result is equivalent to the one you worked out; but they both evaluate to ~ -1.63. Thanks for another great video. I guess I need to file a sympy bug report ...
What you see at first glance heavily depends on your experience and day to day business. There must be a solution between -2 and -1. The solution will not be in Q, which means it will almost for sure have some log(2) and log(3) in it. That there's also a whole number solving this equation is just to fool the less experienced students.
I took the log of both sides and turned it into a quadratic. Then I tried to apply the quadratic formula to it but it just became ugly. I totally forgot I could divide that quadratic by x-1, since x=1 is trivial.
Fabulous. I wanted to do something like this but I didn't succeed. Then, considering that 1 is a solution and transforming the equation to a quadratic equation using natural log, this equation has 2 solutions and I used the relation between coefficients and root to conclude (lets x1 and x2 be the 2 solutions, we have x1=1 and x1*x2 = -ln(6)/ln(2), easy to conclude now)
You don't that path necessarily which requires use of the absurd e^0=1. You can simply take the ln on both sides and solve it like a 2nd degree equation.
Ran Goal Seek in Excel. If the starting guess value is, say x=+2, then the solution converges to x=+1. If the starting x=-2, then it converges to x=-1.63. This is a brute way of quickly getting both the solutions for those mathematically challenged like me.
Surely it would just be easier to tell people to write all the terms as powers of two using logs and then solve the quadratic equation you get when you divide both sides by the term on the right
2^x*3^x =6 right we write 2 =6/3 and 3=6/2 then we get 6^2x= 6*(3^x)*(2^x) write 6 = 3*2. Now 6 ^2x= 3^(x+1)*2^(x+1) 2x=x+1. Therefore x=1. Idk how i figured ts out myself im surprised
"Some might think its completely easy, while others might think its completely impossible. While, the truth is somewhere in the middle" . Yes Its easily impossible ☺️
The way the logarithms and exponents were handled seemed a bit over-complicated to me. I just used logarithms on both sides of the original equation so that it became: (log 3) x^2 + (log 2) x = log 6 = log (3*2) = log 3 + log 2. Replacing the logs with letters and rearranging, it becomes: A x^2 + B x - (A+B) = 0, where A = log 3, B = log 2, and A+B = log 6. That factors as (x - 1) (Ax + (A+B)) = 0. [(A+B)/A = log 6/log 3 = ~1.63] As a result x = 1 or ~-1.63.
Not sure if this helps. Please bear with me. The question asked for "all real values" of x. Because x is a real number, it can be positive, negative, or zero. Zero is not the answer because 2⁰ × 3⁰ ≠ 6, obviously, so there are only two possible scenarios left: x being positive or x being negative. The first assumption only covers the first scenario. The second scenario with x being negative means that 2^x will be less than 1. Since x is still a non-zero real number, x² is definitely positive, and 3^x² will be more than 1. It is then possible to have a negative number x that makes 3^x² larger than 6 while, when multiplied with 2^x (which is less than 1, as mentioned) would yield 2^x (6) that equals to 6.
Suppose you have a quadratic equation which has two solution. The correct way to solve it would be to use the quadratic formula or use other methods that you might have learnt. However, you see by observation that 1 is a solution. But you can't stop there. There still another solution that you haven't found. So you'll have to use the other methods to find the other solution.
@@Zkflames To me ln indicates that the solution has something to do with the number e. While in fact e is nowhere to be found in the solution. I don't like the deception. Also since we can use any base when writing it as a fraction, there are too many ways to write it as a fraction making comparing solutions have a unnecessary extra step.
@@ParkourGrip good point but practically speaking it is easier to calculate the natural logarithm than logarithms with different bases. In fact, if you were to calculate log_3(2) it might even be easier to compute it as ln2/ln3.
Yet another very nice little problem - it is challenging but it can still be solved in the head from the thumbnail without any equipment. I like these a lot, thanks! :) (I did it with the quadratic formula.)
Solution: 2^x*3^(x²) = 6 = 2^1*3^(1²) | The same operations are done with x on the left side of the equation as with 1 on the right side of the equation. Therefore is: x = 1.
It was fairly straightforward to solve using logs resulting in the quadratic (ln3)x² +(ln2)x - ln6 = 0. I then sort of "cheated" as I knew x=1 must be a solution, so (x - 1) must be a factor of the above. This forces the above to factorise into (x - 1)(xln3 + ln6) = 0.... ...which gives the other solution as (-ln6)/(ln3) which is equivalent to your expression.
The answer is pretty simple and nice, just convert whole equation by e^ln(a) and then we need to use quadratic formula to get the answer. 1or -16309... Thanks for that question!
I solved a slightly different way. I turned 6 = 2^x . 3^xx into 6 = 6^x . 3^x(x-1) Then took the log_6 of both sides... 1 = x + log_6(3)x(x-1) Let L = log_6(3) So, we just need to solve the quadratic: Lxx + (1-L)x - 1 = 0 Which has solutions: x = 1 and x = -1/L ~ -1.631
" Some students may think that it is easy, some students may think that it is completely impossible. " Some students may think, some students may not think.
(2^x)(3^x²) = 6 Clearly, x= 1 is a solution. In search of a nontrivial solution: Taking the natural log of both sides gives xln2 + x²ln3 = ln6 = ln2 + ln3 Collecting ln3 and ln2 terms gives (x² - 1)ln3 = (x + 1)(x - 1)ln3 = (1 - x)ln2 Dividing both sides by x - 1 gives (x + 1)ln3 = -ln2 Dividing both sides by ln3 gives x + 1 = - ln2/ln3 Subtracting 1 from both sides yields the second solution x = -(1 + ln2/ln3)
As an aside: Looking *only* at the original problem (and nothing else -in particular without proceeding farther with the actual solution) is it possible to determine how many solutions the equation will have (setting aside degenerate/equivalent solutions)? If so, how would one go about this?
As there is the parabola x^2 in the formula you will expect that there's a positive and a negative solution. As neither -2 nor -1 are a solution, there must be a solution that's not in Q but in R, which leads to the conclusion, that the solution must have some log(2) and log(3) in it. That Presh started with x=1 was just a lucky guess, but sometimes it works 😃.
The best and most natural way to approach this is to take the log with base 6 of both sides, which reduces the problem to a trivial quadratic equation. This problem is basically screaming to be solved in this very way.
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
Seeing that x=1 is a solution is trivial, but solving this equation the proper way sure isn't.
The clue that there is another solution is that there is a degree 2 term (x^2) ...
If you can see an easy solution by inspection, the challenge is always finding the other solutions or proving that there are no other solutions.
Nah, even solving for the general case is trivial for this problem.
As an engineer, y’all are a bunch of silly gooses
triviality in math is the simplest possible solution @@the-boy-who-lived
When you use the zero product property, it is usually wise to check the answer in the original equation. I would have tacked that on to the end of the video.
almost forgot that. yes, it's absolutely necessary.
edit : just checked using a scientific calculator, the 2nd answer is correct.
Every step in this one is logically equivalent to the step above, so here there isn't a need.
@@farhanrejwanIt’s absolutely not necessary since he used equivalences on each step, meaning he can also go back. Or in other words, he also proved that if x=the solution, then x satisfies the equation, at the same time while proving that if x is a solution then x=that solution he found
@@youssefchihab1613 i'm not entirely convinced. i've done such steps before, and the answers didn't satisfy the original question (that one probably had squares and roots, come to think of it).
@@farhanrejwanYeah that'll do it, the usual culprit is when you have x=y and then go to x^2=y^2.
Step 1: Find that 1 is a solution.
Step 2: Imagine a graph of an exponential function to decide whether there is another solution.
Step 3: Watch the video, because you can't solve the equation.
The graph isn’t the graph of an exponential function though, because 3 is raised to the power of x^2.
Step 0: Realize that there is a bracket missing in the equation and that it does not say 3^2x, but 3^(x^2).
Step 1: Fins that 1 is a solution
Step 2: Good job, you found the solution
My inelegant solution involved converting 3 and 6 to 2^(ln3/ln2) and 2^(ln6/ln2) respectively, then an evil quadratic appears, and with a calculator get the answers 1 and -1.631..... Presh's solution is much nicer!
Amazing! How'd you manage that?
@@samuelvieira9748 2^x= 3, ln 2^x = ln 3; x ln 2 = ln 3; x=ln 3/ln 2....
You can divide the quadratic by x-1 to get the other solution, which is -log(6)/log(3)
@samuelvieira9748
I don't know how they did it exactly, but here's what I did, which is the same idea -- except I worked it through to the end, no calculator necessary unless you want the decimal expansion.
2^x * 3^x^2 = 6
Divide by 6
2^x / 2 * 3^x^2 / 3 = 1
2^(x - 1) * 3^(x^2 - 1) = 1
Change of base from base 3 to base 2.
2^(x - 1) * 2^[ln(3^(x^2 - 1))/ln(2)] = 1
n^a * n^b = n^(a + b)
2^[(x - 1) + (x^2 - 1)ln 3 / ln 2] = 1
Apply natural log, and simplify using: ln a^b = b ln a
[(x - 1) + (x^2 - 1)ln 3 / ln 2] ln 2= 0
Distribute and simplify
(x - 1) ln 2 + (x^2 - 1) ln 3 = 0
Factor perfect square.
(x - 1) ln 2 + (x - 1) (x + 1) ln 3 = 0
(x - 1)(ln 2 + [x + 1] ln 3) = 0
By the zero property of multiplication:
x - 1 = 0
x = 1
ln 2 + [x + 1] ln 3 = 0
(x + 1) ln 3 = - ln 2
x + 1 = - ln 2 / ln 3
x = -1 - ln 2 / ln 3
if you do it symoblically you should obtain the same answer
If you haven't learnt about Natural Log (ln), the Normal Base 10 Logarithm (log) can also be used instead.
Couldn't you simplify even further with a change of base log so it would become log_3(2)?
@@bubtb-yl8lu yes
whadda hel you guys talking about(i regret sleeping out my classes)
@@Derryll_Banner Hes saying -Log(6)/Log(3) = -Log₃(6)
because
Logₐ(x) = Logₑ(x)/Logₑ(a) .. Log base change rule
@@whodoesntcare Ok good to know.
When you end by giving an approximate decimal answer, it would be good to also show quick plot on Wolfram Alpha or something.
Got both solutions. Working through it using the quadratic formula, you get to a point where you're questioning reality: "I know that all of this square-root mess needs to simplify to 1 for one of the solutions, but how to get there...?" The key insight for my method was to notice that the property ln(6) = ln(2x3) = ln(2) + ln(3) will lead to a perfect square that will negate the square-root operation.
I did the same thing, except that I (somehow) failed to notice the x=1 solution before I started. But after stumbling across the same trick to simplify the sqrt term, I got (-log3(2) ± (log3(2) + 2)) / 2, and said "oh yeah duh" when one of the solutions was x=1. And then double-checked that the other solution actually worked.
Nice, I missed that perfect square. With that hint I was able to simplify my answer to Presh's form.
quadratic formula ax^2+bx-c=0 a=ln3 b=ln2 c=-ln6
Indeed bro this is exactly what I came down to as well!
Same here ... but it gave a monstrous delta and convoluted x1 and x2 ... didn't seem to me that I could simplify them to get Presh's solituons, was expecting at least one of them to easily turn to 1 !!
@@AZ-yx9bf If you rewrite ln6 as ln2+ln3, the stuff in the square root simplifies to (ln2 + 2ln3)^2 and the answers pop out nicely.
@@AZ-yx9bf ln6=ln2+ln3，D=(ln2)^2+4ln3*(ln2+ln3)=(ln2+2ln3)^2
You don’t even need to apply the quadratic formula. You can divide that quadratic by x-1 since x=1 is a solution.
Man, I just figured "2 × 3 = 6", so X can't be bigger or lower than 1.
I feel like I'm gonna lose marks for not showing extended working...
Same here
Yes, it's the x^2 that messes it up: 3^(x^2) is not an increasing function. If it were then your original reasoning would have been fine.
But you for forgot to consider the negative values. Despite x being negative x^2 is always positive. So in negatives while 2^x decreases, 3^x^2 inxreases, making your argument invalid for the general case.
If the question only asks about positive real values, then your solution is a perfectly acceptable solution.
This question is proof that logarithms were the first sign of fun in math being ruined.
My first trial for any question: x =1?
Most of the time its YES!
Unfortunately, in this case, you would have missed the second solution.
yes it works for finding one of the solution in most cases lol
5:50 Also it equals -log_3(6)
The same as the second answer given, just written differently
After watching the video upto 1:49, I suddenly knew how to approach it. It's very easy take log both sides and solve the quadratic in x.
I knew there would've been an another solution besides x = 1 because of the x^2, but I knew it wouldn't be an integer or anything and I didn't wanna bother. Good video.
See, I just took the natural log of both sides at the start and got xln(2)+x^2 ln(3)=ln(6) converted that to x^2 ln(3)+xln(2)-ln(6)=0 and then used the quadratic formula. I can see your approach being the answer used on a math test where no calculator is available, but that's the route my mind immediately went to.
Actually the quadratic formula also simplifies to (-ln(2) +- (ln(2) + 2ln(3)))/(2ln(3))
that’s what i did too, i’m surprised that’s not how the video went
In all honesty ... yes your solution worked, but you really made a mess of it.
For gods sake ... just take the log on both sides immediately ...
ln(2^x * 3^x^2) = ln6
ln(2^x) + ln(3^x^2) = ln6
x ln2 + x^2 ln3 = ln6
or
x^2 ln3 + x ln2 - ln6 = 0
use quadratic formula and you're done.
Your solution gave you a way to factor it without quadratic formula but that is coincidence ...
just replace the 6 on the right side by 7 and your solution is toast ... while the one shown here
still works.
Or combine the two. Use your method, but also use the fact that at first glance you see that x=1 is a solution. That often helps solve quadratic formulas quicker.
Then you see ln6 is ln3+ln2, you can get the factor (x-1) out and get the other factor in your head without using paper. Filling in the quadratic takes some time.
Calm down. Both are valid approaches.
Agreed. It made no sense to divide both sides by 6, and just complicated the work.
In high school, i had a teacher that would have problems like this on tests. Problems that were almost logically too easy to solve, but would then only award half a point if you didn't show your work.
‘We now have a manageable equation’…🤯
There is a trick with the quadratic, x^2 log3 + x log2 - (log2 + log3) = 0. Since you know one solution is x=1 then the other is always x=c/a= - (log2 + log3)/log3 = -(1 + log2/log3). No need to do the full solution. This comes from c/a = root1*root2 for all quadratics.
I thought this was going to be a gimmicky video about how a^(b^2) is different than (a^b)^2 but i was pleasantly surprised.
Yeah....took me a while too to figure out why that way doesn't work.
Yeah weirdly they decided that a^b^2 = a^(b^2) instead of (a^b)^2 in this problem even though that goes against the standard convention. So I was like oh I see it's just one (although there could be another solution which there is here), but then I dismissed that remembering that a^b^2 is actually (a^b)^2 which make 1 an incorrect answer. Instead you would get to say a^b^2 = a^2b which would completely different than the problem he solved in the video. Suffice to say I find the conventions around exponents and when which ones get applied very confusing. If anyone can explain how you are supposed to see this a^b^2 and know it is a^(b^2) different than other times it is written like this but it would be a^b^2 is treated as (a^b)^2, please let me know.
@brcoutme stop trying to intuit some convention that doesn't exist and watch the video for proper context clues... he explains in words what the equation is actually supposed to be interpreted as. Without parentheses it is just illdefined.
@@brcoutme Standard convention is a^b^c = a^(b^c). For (a^b)^c you have to write brackets, or just a^(bc).
just used a=e^ln(a) and got e^(ln(2)*x + ln(3)*x²)=e^ln(6), and then removed the "e^" part on both side, then just p/q-formula or quadratic.
We can also use quadratic formula.
x^2 log3 + x log2 - (log2 + log3) = 0
x = ( -log2 ± √( (log2)^2 + 4 log3 (log2 + log3) ) / 2 log3 = ( -log2 ± √( (log2)^2 + 4 log3 log2 + 4(log3)^2 ) / 2 log3 = ( -log2 ± ( log2 + 2 log3) ) / 2 log3
So x = 1 or -(1 + log2 / log3)
no thx, i'd rather factor out difference of squares and stuff
That’s what I did
Since you know the other solution is x=1 the other answer is always x=c/a, - (log2 + log3)/log3. No need to do the full sln.
I have no idea what just happened
😂😂😂 I feel your pain
Just take log in both sides and you can solve easily. It seems strange cause the answer ain't simple integer.
😵💫😅
so you don't know logarithms?
Bro knowing loagarithm doesnt mesn that you could directly solve these problems@@ΣΜΡΤΥ
Best method IMO is: spot that x = 1 works. Then use logs from the equation (not the long way like Presh) and get x^2 ln 3 + x ln 2 = ln 6, subtracting ln 6 from both sides to have 0 on the right. Then the knowledge that x - 1 must be a factor helps us factorise it as (x - 1)(x ln 3 + ln 6) = 0.
That's exactly what I did, though I liked Presh's approach of dividing by 6 first.
After taking the log with the correct choice of base, it’s just a quadratic x^2 + (log 2)x - log 6 = 0 which you then just solve by factorising.
Honestly this one was pretty easy, also I am pretty sure there are other ways for solving for both solutions but yours was quite straightforward and simple👍🏼
For a more specific answer, the non-trivial answer is equal to log3(1/6)
How did you drag "taking the log" into a six minute video? Amazing. Such content.
At 3:20 you can also see that we have (2*3^(x+1))^(x-1)=0
Now we see that either x=1(trivial solution), or 2*3^(x+1) is 1(as it is positive)
Then you take log and voila
Log functions, that's all you need, no questions asked
I thought of using logs, but not dividing by 2*3 before doing so. I believe either one results in the solution. Sweet approach.
Solving normally , the discriminant ((log2)^2+4(log3)(log6)) can be factorized to ((log2+2log3)^2). But its way easier to try and rearrange to factor out x-1 as we know one solution. The easiest solution according to me is using that x1+x2 = (-b/a) or x1 * x2 = (c/a) . since x1 = 1 its trivial to obtain x2
After figuring out that x=1 is a solution, we can do the following:
2^x * 3^(x^2) = 6
2^(x-1) * 3^(x^2-1) = 1
x^2-1 is a difference of squares
2^(x-1) * 3^((x+1) * (x-1)) = 1
Since x=1 was already found intuitively, you can raise to the power of 1/(x-1) without worrying about missing a solution
2 * 3^(x+1) = 1
2 * 3 * 3^x = 1
3^x = 1/6
x = log_3(1/6)
x = ln(1/6)/ln(3)
x = -ln(6)/ln(3)
Very clever! I like this approach.
Noice!
Trying to substitute this back in using a phone calculator or various internet calculators is challenging! I think they can't cope with the second exponent.
WolframAlpha plots 2^x 3^x² and provides both solutions for 2^x 3^x² = 6.
Haven't checked other calculators, though.
2=3^a
Ln 2 = a ln 3
a= ln 2 / ln3
(3^ax) × 3^x² = 6
3^(ax+x²)=3^b
6=3^b
b=ln6/ln3
x²+ax=b
x²+ax-b=0
Solve as a quadratic using the formula. Imma use a calc so its easier.
x= 1 first solution
Second solution is -ln6 /ln3.
Ezpz
You can just log both sides straight away and use log rules to make a quadratic ((log(3)x^2 + (log(2)x - log(6) = 0), then use the quadratic equation to solve it for both values (1 and -1.6309…). That seems much easier in my opinion.
(x+1)ln 3 = - ln 2;
ln 3 ^(x+1) = ln 2^(-1)
removing ln from both sides
3^(x+1) = 2^(-1)
3^x * 3 = 1/2
3^x = 1/6
=> x= −1.63093
....
great question!!
I solved this by expanding everything out and using log. Taking the log of both sides got me x^2 log(3) + x log(2) - log(6) = 0. Since we know 1 is a solution, we can factor (x-1) out. This gets us (x-1)(x log(3) + log(6)) = 0, and so x = 1 or x = -log(6)/log(3). Which are the solutions.
Or you can simply take logarithms with base"3" and obtain a very easy equation.
Yup. x = -1 - log_3(2) = - log_3(6) seems a nice way to put it.
You could log directly, log(3)x2 + log(2)x - log(6) = 0 and solve the quadratic.
It becomes a quadratic equation, with logarithmic coefficients so other solution is -(log2/log3)-1
I stumbled upon a nice solution.
We note that 1 is a solution. We also note that the equation can be reduced to a quadratic with leading and linear coefficients log(3) and log(2), respectively. Thus, since one of the solutions is 1, the other must be -1-(log(2)/log(3)), by Vieta's formulae.
PS: log denotes the natural logarithmic function throughout but other bases would work equally well.
I used a modified form of the quadratic equation but couldn't simplify it easily. I didn't want to do it again, so I pushed through, called the unknown square root z, and finally simplified it by using the solution x=1. I was super proud of this creative route until I read that I could've just used polynomial division and didn't need any quadratic formula at all, which I gracefully forgot about. Sometimes, you can't see the big picture when you're so focused on the details
Very nice!
I did the quadratic equation, then quickly used Horner's scheme to divide (x-1) out, and found the other linear factor to be (ln3 • x + ln6), so x = -ln6/ln3 = -1 - ln2/ln3. 🤓
I did ln(2^x * 3^x^2)=ln(2^x) + ln(3^x^2)=ln(3)x^2 + ln(2)x
Then you do a quick rearrange to get ln(3) as a, ln(2) as b, and -ln(6) as c in the quadratic equation.
Then you use wolfram alpha to tell you that the big messy +- quadratic equation is equal to 1 and -1.63
ln2 / ln3 is also the same thing as log(base3) of 2.
Can't you take the log of both sides?
log(2^x * 3^(x^2)) = log(2^x) + log(3^{x^2)) = xlog2 + x^2log3 = log6
Let b = log2, a = log3, c = log6 and this becomes
bx + ax^2 = c
and then it's just a quadratic equation that can be solved using the quadratic formula?
Am I missing something?
Yes, you can do that. That's how I did it. However, you'll see that you'll get a nasty determinant: √[(ln2)² + 4 ln3 ln6]
It wasn't immediately obvious to me how to proceed her but I figured it out after a while.
You'll have to rewrite ln6 as ln2 + ln3, and then proceed as follows (I'll only look at the argument under the root for simplicity's sake):
(ln2)² + 4 ln3 ln6 =
(ln2)² + 4 ln3 (ln2 + ln3) =
(ln2)² + 4 ln2 ⋅ ln3 + 4 (ln3)² =
(ln2)² + 2 ln2 ⋅ 2 ln3 + (2 ln3)²
Now, we have a perfect square of ln2 and 2 ln3. 😀
(ln2)² + 2 ln2 ⋅ 2 ln3 + (2 ln3)² =
(ln2 + 2 ln3)²
Applying the square root obviously only leaves us with ln2 + 2ln3.
After that, the two solutions of the quadratic formula collapse to either 1 or -(ln2 + ln3) / ln3 which can also be written as -ln6 / ln3.
Knowing that things get a little tricky with the quadratic formula, Presh's solution turns out to be the easier approach, in my opinion, but I must admit that it hadn't occurred to me to do the steps he did in the beginning (i.e. dividing both sides by 6, splitting 6 into 2⋅3 and then work these factors into the exponents of the numerator).
Better still, if you choose the correct base (when you can choose the value of one coefficient, which value do you choose of which coefficient?) then it factorises straight away.
x^2 + (log 2)x - log 6 = 0
x^2 + (log 6 - 1)x - log 6 = 0
(x + log 6)(x - 1) = 0
x = -log 6, x = 1
change of base formula into quadratic seems easier
if you have a calculator you can just take the natural log of both sides, use properties of logs to split 2^x and 3^x^2 into addition, use more properties of ln to get the exponents to the front of the natural logs, subtract over and use the quadratic equation to get the solution immediately without all these hoops of logic.
Every time I watch these videos I come out happier.
I’d have just written both of the bases as e^ln(base), added the exponents together, taken the ln of both sides, and gotten a quadratic that you can probably just quad formula.
Thanks from Canberra 🇦🇺 Tesh.
I used the quadratic equation instead of removing the factors of 2 and 3 from the RHS, getting x=1 and x=\frac{-\ln2-\sqrt{(\ln2)^2+4\ln3\ln6}}{2\ln3}
This was a good one. I used sympy to try to solve this equation but it threw up it's hands, giving this error:
NotImplementedError: multiple generators [2**x, 3**(x**2)]
No algorithms are implemented to solve equation 2**x*3**(x**2) - 6
I then used Wolfram Alpha and it gave x = {1, - log(6) / log(3) }. It is not immediately obvious how this result is equivalent to the one you worked out; but they both evaluate to ~ -1.63.
Thanks for another great video. I guess I need to file a sympy bug report ...
In the second approach logarithmic equation gives the approximate value to the x which give 5.98.....=6
Interesting and elegant!
Good one 👍
I'm far too rusty at math. 😅 My first instinct was, "can I do some log wizardry?" but I clearly did not do it *correctly*.
Nice. Good reminder too of a few logarithm tools.
What you see at first glance heavily depends on your experience and day to day business. There must be a solution between -2 and -1. The solution will not be in Q, which means it will almost for sure have some log(2) and log(3) in it. That there's also a whole number solving this equation is just to fool the less experienced students.
I took the log of both sides and turned it into a quadratic. Then I tried to apply the quadratic formula to it but it just became ugly. I totally forgot I could divide that quadratic by x-1, since x=1 is trivial.
Fabulous.
I wanted to do something like this but I didn't succeed.
Then, considering that 1 is a solution and transforming the equation to a quadratic equation using natural log, this equation has 2 solutions and I used the relation between coefficients and root to conclude (lets x1 and x2 be the 2 solutions, we have x1=1 and x1*x2 = -ln(6)/ln(2), easy to conclude now)
Nice! Just how I did it.
You don't that path necessarily which requires use of the absurd e^0=1. You can simply take the ln on both sides and solve it like a 2nd degree equation.
Congratulations on reaching pi million subscribers!🎉
2x×3x² = 6 ; 6x³ = 6 ( 6 is cancelled), 1x³= 1 therefore x³=1 if x^3 = 1 the x = cube root 1,which is 1 so x=1
edit .
That was insane 😅. II enjoyed the video. Greetings from Peru
Ran Goal Seek in Excel. If the starting guess value is, say x=+2, then the solution converges to x=+1. If the starting x=-2, then it converges to x=-1.63. This is a brute way of quickly getting both the solutions for those mathematically challenged like me.
Surely it would just be easier to tell people to write all the terms as powers of two using logs and then solve the quadratic equation you get when you divide both sides by the term on the right
I might be wrong but couldn't 3^(x²) can be written as 3^(2x)???
No, there is no such rule
What you mean is (3^x)^2=3^(2x).
but x is just negative of golden ratio. That could've been done, after noting that the qudratic equation for x is: x^2 + x - 1 = 0.
Sir why not , 2^x.3^x.3^x =(18)^x. Then take log and solve it ??
2^x*3^x =6 right we write 2 =6/3 and 3=6/2 then we get 6^2x= 6*(3^x)*(2^x) write 6 = 3*2. Now 6 ^2x= 3^(x+1)*2^(x+1) 2x=x+1. Therefore x=1. Idk how i figured ts out myself im surprised
"Some might think its completely easy, while others might think its completely impossible. While, the truth is somewhere in the middle"
.
Yes Its easily impossible ☺️
😂
.... or impossibly easy! 😛
The way the logarithms and exponents were handled seemed a bit over-complicated to me.
I just used logarithms on both sides of the original equation so that it became:
(log 3) x^2 + (log 2) x = log 6 = log (3*2) = log 3 + log 2.
Replacing the logs with letters and rearranging, it becomes:
A x^2 + B x - (A+B) = 0, where A = log 3, B = log 2, and A+B = log 6.
That factors as (x - 1) (Ax + (A+B)) = 0.
[(A+B)/A = log 6/log 3 = ~1.63]
As a result x = 1 or ~-1.63.
can you explain why the first way doesn't give us all the solutions?
Not sure if this helps. Please bear with me.
The question asked for "all real values" of x. Because x is a real number, it can be positive, negative, or zero. Zero is not the answer because 2⁰ × 3⁰ ≠ 6, obviously, so there are only two possible scenarios left: x being positive or x being negative. The first assumption only covers the first scenario. The second scenario with x being negative means that 2^x will be less than 1. Since x is still a non-zero real number, x² is definitely positive, and 3^x² will be more than 1. It is then possible to have a negative number x that makes 3^x² larger than 6 while, when multiplied with 2^x (which is less than 1, as mentioned) would yield 2^x (6) that equals to 6.
Suppose you have a quadratic equation which has two solution. The correct way to solve it would be to use the quadratic formula or use other methods that you might have learnt. However, you see by observation that 1 is a solution. But you can't stop there. There still another solution that you haven't found. So you'll have to use the other methods to find the other solution.
Simplify solution notation: ln(2) / ln(3) = log3(2)
Don't write it that way. The base is to be a subscript: log_3(2)
that is just another way of writing it. not simplified... I would consider ln2/ln3 to be more simple in my opinion
@@Zkflames To me ln indicates that the solution has something to do with the number e. While in fact e is nowhere to be found in the solution. I don't like the deception. Also since we can use any base when writing it as a fraction, there are too many ways to write it as a fraction making comparing solutions have a unnecessary extra step.
@@ParkourGrip good point but practically speaking it is easier to calculate the natural logarithm than logarithms with different bases. In fact, if you were to calculate log_3(2) it might even be easier to compute it as ln2/ln3.
But it says 2^x*3^x^2=2^x*3^(2x)=6. There is a bracket missing in the equation.
What is the result for this equation?
Good god finally something that looks intimidating in this channel
Yet another very nice little problem - it is challenging but it can still be solved in the head from the thumbnail without any equipment. I like these a lot, thanks! :)
(I did it with the quadratic formula.)
Solution:
2^x*3^(x²) = 6 = 2^1*3^(1²) | The same operations are done with x on the left side of the equation as with 1 on the right side of the equation. Therefore is: x = 1.
It was fairly straightforward to solve using logs resulting in the quadratic (ln3)x² +(ln2)x - ln6 = 0.
I then sort of "cheated" as I knew x=1 must be a solution, so (x - 1) must be a factor of the above.
This forces the above to factorise into (x - 1)(xln3 + ln6) = 0....
...which gives the other solution as (-ln6)/(ln3) which is equivalent to your expression.
Awesome problem. Just a little disappointed there wasn’t a simplification of putting the second solution back into the original equatiin
Beautiful solution!!! Like watching a magic trick.
The answer is pretty simple and nice, just convert whole equation by e^ln(a) and then we need to use quadratic formula to get the answer. 1or -16309... Thanks for that question!
You could simply take logs, shift the ln(6) top the other side and solve using the quadratic formula?
I solved a slightly different way.
I turned
6 = 2^x . 3^xx
into
6 = 6^x . 3^x(x-1)
Then took the log_6 of both sides...
1 = x + log_6(3)x(x-1)
Let L = log_6(3)
So, we just need to solve the quadratic:
Lxx + (1-L)x - 1 = 0
Which has solutions:
x = 1
and
x = -1/L ~ -1.631
" Some students may think that it is easy, some students may think that it is completely impossible. "
Some students may think, some students may not think.
(2^x)(3^x²) = 6
Clearly, x= 1 is a solution. In search of a nontrivial solution:
Taking the natural log of both sides gives
xln2 + x²ln3 = ln6 = ln2 + ln3
Collecting ln3 and ln2 terms gives
(x² - 1)ln3 = (x + 1)(x - 1)ln3 = (1 - x)ln2
Dividing both sides by x - 1 gives
(x + 1)ln3 = -ln2
Dividing both sides by ln3 gives
x + 1 = - ln2/ln3
Subtracting 1 from both sides yields the second solution
x = -(1 + ln2/ln3)
How is 1 a valid solution? If we calculate with x=1, we get 18.
3^x^2 = 3^(x*2) => 9
I would have been tempted to start with (2^x)(3^x)(3^x) = (3^x)(6^x) = 6 but then I remembered that not how exponents work.
As an aside:
Looking *only* at the original problem (and nothing else -in particular without proceeding farther with the actual solution) is it possible to determine how many solutions the equation will have (setting aside degenerate/equivalent solutions)?
If so, how would one go about this?
(1) notice 1 is a solution (2) derive that pesky quadratic (3) as taught in 10th grade factor out by eye: (x-1)*(x*ln(3)+ln(6)) done.
I can't believe it. I actually managed to get this one right before watching the solution
is x the golden ratio?
By inspection, x = 1
Yeah, that's the easy part. If only it didn't ask for *all* solutions! 😅
Is there any reason why one solution is obvious to find and other not?
As there is the parabola x^2 in the formula you will expect that there's a positive and a negative solution. As neither -2 nor -1 are a solution, there must be a solution that's not in Q but in R, which leads to the conclusion, that the solution must have some log(2) and log(3) in it. That Presh started with x=1 was just a lucky guess, but sometimes it works 😃.
Nice question ❤❤❤❤❤❤❤
The best and most natural way to approach this is to take the log with base 6 of both sides, which reduces the problem to a trivial quadratic equation. This problem is basically screaming to be solved in this very way.
Thanks Master 🙏🙏
Ngl i use trail and error method most of the time for these type of questions 🤣🤣🤣
This often works in problems that have been designed to have "nice" solutions.
The problems start when one is supposed to provide *all* solutions! 😉