Using a logarithm property to make this equation easier!
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- Опубликовано: 4 июл 2024
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We will solve this power equation x^ln(4)+x^ln(10)=x^ln(25). However, the process of solving this equation would require making one side equal to 0 but factoring every term by x^ln(4), and I don't think the expressions will look that nice. So instead, we can use the log property that a^logb(c)=c^logb(a) and change the equation to an exponential equation. The answer will involve the usage of the quadratic formula and we will also see the golden ratio.
Equation-of-the-year playlist: • Equation Of The Year!
0:00 Math for fun, solving the power equation x^ln(4)+x^ln(10)=x^ln(25)
1:09 Proving a^logb(c)=c^logb(a)
3:18 Solving the equation in exponential form
8:35 Check out Brilliant
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Your ability to change colors quickly doesn't cease to amaze me.
Fr. So seamless
also erasing half the board with a tap is a nice skill
Average asian mathematics lecture's muscle memory
😂😂@@gubunki
@@nut4ku wait hes asain??
I tried it, but came up with a different method:
x^ln4 + x^ln10 = x^ln25
a = x^ln5 , b = x^ln2
b^2 + ab = a^2
Using Quadratic Formula:
a = b(1 + root5)/2 [since a and b must be the same sign, the other one is ignored]
x^ln5 = x^ln2(phi) [dividing by x^ln2, means x^ln2 must not be 0. So 0 is a Solution. 😅 Didn't realize at first.]
x^ln(5/2) = phi
x = phi^(1/ln(5/2))
Edit: Essentially I found out after watching the video for his solution that phi^(1/(ln(5/2)) = phi^(log(base5/2)(e)) = e^log(base(5/2))(phi). So I got the Answer!!! Woohoo!!
You should divide both side by b^2. And then use quadratic formulas with
(A/b)^2+(a/b)-1=0
Both end up with the same relation a and b. b comes out common anyway for the relation. If I had divided by b^2, I would have gotten a/b = phi. It is just my instinct to use it without dividing, as I had practiced all along.
That was my method too!
I did it this way too (except I made b=x^ln5 and a=x^ln2). Got phi^(1/(ln5-ln2)) which is the same thing.
sameee
The golden ratio caught me off guard, but the fact that it’s there is amazing
Been a long time since I have seen a math equation and problem that was just genuinely cool. Having phi show up was a real nice surprise.
Seems like we're finally back at the golden ratio we all love so much - one BPRP video a week.
He’s posting daily on bprp math basics
@@memebaltan Herzlichen Dank!
2:50 Voodoo
what i did was move everything to one side and factor x^ln4(x^ln25/4 - x^ln10/4 -- 1) = 0 --> x^ln25/4 is just x^ln10/4 squared so in this case u can let t = x^ln10/4
leading to the quadratic equation t^2 - t - 1 = 0 --> t solves for (1+-sqrt(5) / 2 --> x^ln10/4 = (1+--sqrt(5) / 2 --> take ln() on both sides --> ln10/4*lnx = ln((1+--sqrt(5)/2)) -->
lnx = ln((1+--sqrt(5)/2)) / ln10 --> to solve for x, e^ on both sides to cancel the lnx to get x --> x = e^(ln((1+--sqrt(5)/2)) / ln10) --> x ≈ 1.69
This was my idea:
x^ln4 + x^ln10 = x^ln25
Leave out the case x=0 and divide both sides by x^ln10. Distributing the denominator and applying properties of the power you get
x^(ln4-ln10) + 1 = x^(ln25 - ln10)
Applying properties of logarithms we get
x^ln(2/5) + 1 = x^ln(5/2)
Substituting z=x^ln(5/2), and noting that ln(2/5)=-ln(5/2) the equation becomes
z^-1 + 1 = z
Which can easily become a quadratic equation and the rest follows in a similar way to what you did.
Wolframalpha couldn't figure out the given exact form solution from the original form, but could when the equation was transformed.
Got it in 2 min buddy!!
You definitely helped me a lot to prepare logarithms for competitions..!
Gratitude from India 🇮🇳
Answer is nearly equal to 1 + ln(2)
You're a phisic, aren't you?
You can also start by dividing everything by x^ln(4) and immediately getting a quadratic of x^ln(5/2). Nice equation.
True!
Fiz desta forma.
Exactly what I did
Yes. I took it out as a factor.
very nice equation and explanation
Very nice equation. Thanks for your perfect presentation. Great, as always👌your videos are so great!
FINALLY!! A youtube math exercise that I couldn't do in my head. Nice. It brought back some memories of forgotten log properties. Oh, to be a freshman in college again.
I dont understand half of the video but it's still fun to watch
Can we just acknowledge the shear number of markers in the bottom right bro be spending 25 hours a day on math to use them up
Sheer
He needs an expo sponsorship
@@erichury indeed
Try to solve this question that I found in a calculus textbook (by James Stewart):
Use a quadruple integral to find the hypervolume enclosed by the hypersphere x^2+y^2+z^2+w^2=r^2 in R4 (I wish I made that up lol)
😮
Love it, thanks.
Great solution development. I never would have thought to approach it that way, because I never thought about the fact a^ln(b) = b^ln(a). I mean, it's obviously true, but not a tool I would have thought to use.
Very good video!👏👏
Anhother way to isolate the X in the last part would be using the NOTE again, and we get X=(phi)^(1/ln(5/2))
x=φ^(1/ln(²/⁵)) ?
That's a nice solution, ngl
I got the same thing. X=(phi)^(1/ln(2.5)). Most people don't have a "log to the base 2.5" button on their calculator, so getting rid of log to base 2.5 simplifies to something one can plug into a calculator, where phi is the golden ratio constant, which is equal to [(1+sqrt(5)]/2.
from Morocco all my respects...thank you for those genious ideas..i shared this video on facebook
I love every video that starts with Let's do some math for fun. Thanks!
Hey! Have there always been a warehouse full of colored markers in your teaching studio? I've watched all your videos and have never noticed that feature.
Either way, nice touch, like all you do.
That is a really cool problem and solution.
I noticed immediately that it would form a quadratic equation when we divide through either of X^ln4 and X^ln25
But I didn't know about the switch of X and the argument of the log functions.
I really love your videos thanks so much.
Finally! That is indeed a nice one!!
I gave this baby a try, and was surprised I could get the answer (got an imaginary one as well), but I did this a little differently. I noticed the common factors 4 = 2*2, 10 = 2*5, and 25 = 5*5, so I knew there was some way I could leverage that, so I factored it, and with some playing around I got: x^2 ln 2 + x^ln 5* x^ln 2 - x^2 ln 5 = 0. At this point, I noticed that we had (x^ln 2)^2 in the first term, and x^ln 2 in the second, so used the quadratic formula (solving was interesting but too long for RUclips comment) to solve for x^ln 2, which gives x^ln 2 = x^ln 5 (-1 +/- sqrt(5))/2. Then, I moved the x^ln 5 to the other side and combine to get x^ln(2/5) = (-1 +/- sqrt(5))/2. Now I just take the weird root of both sides and I get the same answer.
You’re lying you didn’t do that. Don’t make up stuff on the internet. Your story doesn’t make any sense. It’s actually insane. No sane person could believe it. There is no way you figured it out and there’s nothing you could say that would make the internet believe you. “Oh, but I explained it,” you’ll say. And I’ll say, “No, I don’t accept your explanation and I think you’re lying.”
Who’s with me??! Let’s let this guy hear it!! We don’t believe a word he’s saying!
@@n16161 that's up to you, bubs. Really makes no difference to me.
This guy is always sponsored by brilliant. He probably has a free course by now from them 😂
I love learning from him. He’s the whole reason I was so interested in integrals
in chess there is something called coursera where many of the top players make courses. (e.g. on openings or whatever) . @fourthofno9184 I don't know how brilliant works but your comment made me think, if blackpenredpen could do a course for them.
I first converted all the ln() functions from:
ln(4), ln(10), ln(25)
to
2ln(2), ln(2) + ln(5), 2ln(5)
Then I divided everybody by x^2ln(5), getting:
x^(2ln(2) - 2ln(5))
+ x^(ln(2) + ln(5) - 2ln(5))
=
1
Which becomes:
u^2 + u - 1 = 0,
where u = x^(ln(2) - ln(5)) = x^(ln(2/5))
though my final solution does not look as elegant:
x = ((-1 + √5) / 2) ^ (1 / ln(2/5))
Pretty cool to see how many ways there are of solving this equation!
Yes, cool solution
Very cool problem.
Love your videos
wonderful!
Nicely done. ❤
As a side note, basically the same approach also works without turning it into an exponential equation, after excluding the solution x=0 just divide the original equation by x^(log 4) on both sides, which gives:
1 + x^log(5/2) = x^log(25/4) = x^(log(5/2) + log(5/2)) = (x^log(5/2))^2
Using substitution u = x^log(5/2) gives 1 + u = u^2, and therefor u = phi. By solving for x we then get the same result
I love how you are able to bring golden ratio everywhere!
4 10 and 25 chosen deliberately for that; it's not an accident
Thanks
I tried to do it in another way by manipulating around the powers but only got to 0, how do I know if an equation like this one has more than 1 solution ? and is there a methodology I could follow to find these solutions? or do I just have to study extremely hard maths to become able to find them
This equation may be solved without using the transformation: x^ln(a) = a^ln(x) however, it makes it more convenient. Very interesting clip and nice explanation.
You can avoid using fractional bases by reversing the exponent a second time near the end:
x^(ln(5/2)) = phi
So x = phi^(1/ln(5/2)) = 1.691...
Blackpenredpen just casually fit the golden ratio and 69 in one equation
Awesome! For the solution, why is e^log(base 5/2)(phi) preferable to phi^(1/(log(5/2)), or phi^log(base 5/2)(e)? ❤
You switch colors just as smoothly as you switch log bases 😄. And that stock of pens in corner..😄
I solved it using only power properties, but I got phi^ln(2/5), but then realized that, doing some additional operations, I got the same result as in the video.
I think it should be phi^(1/ln(5/2))
Thankyou sir,this is goos way.
wow that's cool
when i first saw the question in the thumbnail I thought of relating it to the technique of solving integrals by using log there I used the same way
x^ln(4)+x^ln(10)=x^ln(25)
taking log on both sides
lnx^ln(4)+lnx^ln(10)=lnx^ln(25)
by ln property lnx^a = a ln x
ln(4)*ln x+ln(10)*ln x=ln(25)*ln x
canceling out ln x from both sides....we get
ln(4)+ln(10)=ln(25)
and after reaching here I was like what do we have to find in this in the first place!!🥲
nice and very easy
Difference of two squares and then divide each factor by x^(ln2) giving 2=x^(ln5)
brilliant!
The "equation of the year" definition reminds me when I participated to the local math contest which I participated when I was 12
With a different procedure I also found two complex conjugate solutions approximately x=-0.567 ± 0.167·i
Very good. Cheguei nessa resposta equivalente: [ (sqrt(5-1)/2 ] ^ (1 / ln(2/5) ).
Can u help solve Integral of 1/(Square root of x(to the power 3)+x)
How did you know to do the step at 5:47 where you set one side equal to the quadratic formula value
You can write x^ln4=x^ln(2*2)=x^(ln2+ln2)=x^ln2*x^ln2. Doing soemthing similar for the other terms one finds: x^ln2*x^ln2+x^ln2*x^ln5=x^ln5*x^ln5
Dividing everything by ln2*ln5
x^ln2/x^ln5 + 1 = x^ln5/x^ln2
Define a=x^ln2/x^ln5
Then
a+1=1/a
Which yields a=phi
Then x=phi^(ln5/ln2)=phi^ln(5/2)
Which is what you got expressed differently
I also did the same method, and I posted my take but I actually got phi^(1/ln(5/2)). If a + 1 = 1/a, U get a = ((-1 +- root(5))/2). Which is not phi. U can't take negative. so it will be 1/phi. and answers will match.
@@pranavrs184 aahhhh you are complitely right my bad!!
:)
For the first time in a while I impressed myself. I looked at the exponent, told myself, 4=2*2, 25=5*5 and 10=2*5 that would be nice to divide either by 4 to have a 5/2 and (5/2)**2 appear. Then I looked at the signs in front of the coefficient and I said to myself : I don’t know how but there is the golden ratio hidden in this equation.
So I looked at the nonzero solution to x^(ln(9)) + x^(ln(12)) = x^(ln(16)) and noticed the solution was e^(log base 4/3 (phi)). So this leads to the idea that maybe for positive values a and b the nonzero solution to x^(ln(a^2)) + x^(ln(ab)) = x^(ln(b^2)) is e^(logbase b/a (phi)). I'm going to work this out to see if it is true.
the first transformation isn't material. you can just divide by x^log(25) and simplify: x^(2log(2/5))+x(log(2/5))=1 and you're on the same line.
I was thinking of x=phi^(1/ln2.5). Is there a preferable method of writing it down? And if so, why?
i tried solving it on my own and got spooked by the golden ratio jumpscare
this is brilliant.
If THIS doesn't smell like a hidden quadratic.
Later Edit: And it was. And I haven't even seen the video yet.
I like the stash of markers
you are the best
Solution:
x^ln4 + x^ln10 = x^ln25
x^(2ln2) + x^(ln2+ln5) = x^(2ln5)
(x^ln2)² + x^ln2 * x^ln5 = (x^ln5)² |-(x^ln5)²
(x^ln2)² + x^ln2 * x^ln5 - (x^ln5)² = 0
Substitution
u = x^ln2
v = x^ln5
u² + vu - v² = 0
u = -v/2 ± √((v/2)² - (-v²))
u = -v/2 ± √(v²/4 + 4v²/4)
u = -v/2 ± √(5v²/4)
u = -v/2 ± √5 * v/2
u = v * (-1 ± √5)/2
Resubstitution
x^ln2 = x^ln5 * (-1 ± √5)/2 |ln
ln2 * lnx = ln5 * lnx + ln((-1 ± √5)/2) |-(ln5 * lnx)
ln2 * lnx - ln5 * lnx = ln((-1 ± √5)/2)
lnx * (ln2 - ln5) = ln((-1 ± √5)/2) |:(ln2 - ln5)
lnx = ln((-1 ± √5)/2) / (ln2 - ln5) |e → e^(lna/b) = a^(1/b)
x = ((-1 ± √5)/2)^(1/ (ln2 - ln5))
x₁ ≅ 1.69075...
x₂ ≅ -0.567... + i * 0.167...
Curious, that x = 0 doesn't come up as a result, even though it is clearly a valid solution
What's cool is that e≈5/2, so the log with base 5/2 is approximately ln, so the solution is approximately the golden ratio.
Nice equation❤
That's a lot of pens you got there
I used the same property but then used graphs to find the no of soln
x^(ln 2 + ln 2) + x^(ln 2 + ln 5) == x^(ln5+ln5)
divide by x^(ln2+ln5):
x^(ln2)/x^(ln5)+1==x^(ln5-ln2)
substitute:
t=x^(ln5)/x^(ln2)
t^2-t-1==0
following steps are same
Starting with: x^(ln 4) + x^(ln 10) = x^(ln 25)
Divide both sides by the RHS: x^(ln 4) / x^(ln 25) + x^(ln 10) / x^(ln 25) = 1
Use law of indices: x^(ln 4 - ln 25) + x^(ln 10 - ln 25) = 1
Use law of logs: x^(ln 4/25) + x^(ln 10/25) = 1
Form a quadratic: (x^(ln 2/5))^2 + x^(ln 2/5) - 1 = 0
Solve quadratic: x^(ln 2/5) = (-1 +/- sqrt(5)) / 2
Since powers are always positive, choose + solution only: x^(ln 2/5) = (-1 + sqrt 5) / 2
Therefore x = ((-1 + sqrt 5) / 2)^(1 / ln 2/5)
= 1.69075...
Getting my answer to match the form in the video was the hardest part!
let phi = (1 + sqrt 5) / 2, then 1/phi = (-1 + sqrt 5) / 2
Then x = (1/phi)^(1 / ln 2/5)
x = phi^(-1 / ln 2/5)
x = phi^(1 / ln 5/2)
x = phi^(ln e / ln 5/2)
x = e^(ln e / ln 5/2 * ln phi)
x = e^(ln phi / ln 5/2 * ln e)
x = e^(log_{5/2}(phi) * ln e)
x = e^(log_{5/2)(phi))
I saw the thumbnail and tried solving it myself, And I got the answer (5/2)√[(1+√5)/2] as X.
Always the same trick with those:
divide the 2 sides by smth clever so you end up with a ratio and its inverse.
The 1st step is to decompose the exponents with the ln rules
ln4=ln2+ln2, etc...
Divide both sides of the equation by x^ln2*x^ln5, and you get to 1/X+1=X with X =x^ln2.5
X = phi (well known golden ratio), the negative root is rejected given X>0
x = Exp(lnPhi / ln(2.5))
year 11 stuff.
Please try to integrate 1/x^5+1
does anyone know the promo code (if there's any) for his merch?
Can you please help me with this question sir, x^2=y^2 by taking square root we have
√x^2=√y^2
|x|=|y| what is the next step sir
I got ((1+sqr5)/2)^(1/ln(5/2)), so basically the same thing :D
I believe you can tidy this up a bit by using the change of basis formula of the logarithm. As in:
log(φ,5/2)=ln(φ)/ln(5/2)
Thus, x=[exp(ln(φ))]^(1/ln(5/2))
x=φ^(1/ln(5/2))
Edit: Your solution might actually be prettier lol.
x=e^a, 4^a + 10^a = 25^a
s=2^a, t=5^a
s^2+st-t^2=0
Hmmm, can solve s as a function of t or vice versa as a quadratic, maybe the equation is solvable for a when you substitute back. Perhaps a log or W function will show up later if the problem is nice enough to allow it.
And then I start watching and basically 5/2 and its square are nicely there already, to the a power. I wonder how it goes if I follow through on what I was going to do.
Equivalently, divide by t^2, and s/t=w. Or do it the other was around with s^2.
@@mtaur4113 Try to factor s^2 + st - t^2 by imagining it in its factored form with injected variables like this:
(s + ut)(s - vt) = 0
We can expand this to get something similar to what we started with:
s^2 + (u - v)*st - uv*t^2 = 0
If we want this new formula to end up just like our original formula, then both 'u' and 'v' must have values which satisfy these two equations:
1) uv = 1 (comes from uv*t^2)
2) u - v = 1 (comes from (u - v)*st)
Solving these two simultaneous equations yields us with:
u = (1 + sqrt(5)) / 2
v = (-1 + sqrt(5)) / 2
We can rewrite this with the golden ratio 'phi':
u = phi
v = 1 / phi
Now we can substitute 'u' and 'v' back into our factored formula:
(s + (phi)*t)(s - (1 / phi)*t) = 0
My thumbs are getting tired typing all this on my phone lol, so I'll stop here. But just substitute back in 2^a and 5^a for 's' and 't', and you should be able to solve for a.
@@dannydewario1550 Kind of nice, probably was better just to observe the 5/2 and (5/2)^2 in the first place, but why quit halfway if it's doable?
@@mtaur4113 Exactly!
Plus it was just nice to see someone else in the comments who approached it similar to how I did. I also didn't think about using the quadratic formula with 5/2. This definitely took more steps than the solution in the video, but it's cool that there's more than one method of solving.
Whenever (•)² = (•) + 1 appears, always expect ±φ^(±1).
If bro starts yapping about logarithms again, I'm going to sle
Wait, but why is the golden ratio (1 + sqrt(5)) / 2? that's such a random number!
Beautiful
That's just brilliant! Literaly lol
Im in 9th grade, and it's funny how i understood some of it. That's why because he explained in a way that my sped mind could even understand. And today, i thought myself the Pythagorean Theorem
x=[(√5+1)/2]^(1/ln5/2)=1,69075
I got (phi-1)^(1/(ln(2/5))), which is same.
Nice
If we want to find all the real solutions, we have to check negative numbers as well, but ln x has no real solution if x is negative, so using this way cannot give us negative solutions. My question: is there a way to prove that this equation has no negative solutions or is there a condition for x being positive? Thanks!
genial
Can x be equal to 1 as well?
No, 1 to the power of something is always one, you would get 1+1=1 which is false
I did it differently. x^(2ln2) + x^ln2 x^ln5 = x^(2ln5)
Complete the square on LHS
(x^ln2 + 1/2 x^ln5)^2 - 1/4 x^ln5 = x^(2ln5)
x^ln2 + 1/2 x^ln5 = sqrt(5)/2 x^ln5
Solve for x^ln2, take ln of both sides and with algebra solve for x
x= e^[ln((sqrt(5)-1)/2) / ln(2/5)] ends up being equivalent to BPRP's answer
x is also equal to 1 if you want another trivial solution. Otherwise it's 1.69075256401782556972357870922404325917589445848038.
i love math of course
Hey, I did it!
Why was the complex solution ignored?
You could go further by using formula log(a)(b)=lnb/lna.
So it would be something like x=q-(5/2) [counting in my head..... Someone double check...]
yea there really is no need to use it in this case cuz ppl usually only use it to simplify equations you still want to solve
Disagree. When you look are Bprp final solution there is really no clue what is actually the value of that expression. On the other hand the simplified version is obvious at first sight: "a little bit more than -1" 😁
Doesn't your first step exclude complex solutions?