I went straight in with a similar method to your #2. As the LHS is a factor of the expression in the RHS I put u = 2^x - 3^x And squared both sides to give a quadratic in x and we get that log expression as the other solution.
Before watching the video: Given: 2↑x - 3↑x = √(6↑x - 9↑x) To find: x Separating RHS into prime factors to match LHS: 2↑x - 3↑x = √(2↑x·3↑x - 3↑2x) Taking 2↑x as a and 3↑x as b: a - b = √(ab - b²) Squaring both sides: a² - 2ab + b² = ab - b² a² - 3ab + 2b² = 0 a² - 2ab - ab + 2b² = 0 a(a - 2b) - b(a - 2b) = 0 (a - b)(a - 2b) = 0 from here, a = b, or a = 2b. a = b → 2↑x = 3↑x Taking log of both sides: xlog2 = xlog3 Moving all terms to LHS: xlog2 - xlog3 = 0 x(log2 - log3) = 0 As log2 cannot be equal to log3, x = 0. a = 2b → 2↑x = 2·(3↑x) Taking log on both sides: xlog2 = log(2·(3↑x) Using logab = log a + logb: xlog2 = log2 + xlog3 Taking all x terms in LHS: xlog2 - xlog3 = log2 x(log2 - log3) = log2 xlog⅔ = log2 x = log2/log⅔. Testing, x = 0 trivially works as it reduces both sides to 0. When x = log2/log⅔, I'm literally not capable of testing.
Alternative solution: a - b = √(ab - b²) a - b = (√b)√(a - b) (a - b) - (√b)√(a - b) = 0 (√(a - b))(√(a - b) - √b) = 0 a = b or √(a - b) = √b → a = 2b Proceed as above.
My method is similar to your first method, except that I solved for a in terms of b directly after getting the quadratic. From there, I got the correct solutions.
Nice one! I have become a HUGE fan of your Math and maybe more.. the below comments are for the "more" part (just on a lighter note, hope you don't mind) 1) You should be awarded a Master's degree - Master of Substitutions 2) You have the knack of somehow bringing in your "2b or not 2b", "Let's call it 'y'; now don't ask me 'why'" puns... So maybe a 'pun'dit of Substitutions 3) I waited till the end of this video - felt almost like waiting for the post-credit scene after a Marvel movie
I didn't start with a substitution. Squared both sides first and distributed the LHS. After distribution you get 4, 6 and 9 as the bases. Moved the whole RHS to the left. Which gives you 4^x-3*6^x+2*9^x=0. Now I divided both sides by 4^x (as 4^x can't be zero) and then substituted (3/2)^x as Z. Which gives you a quadratic equation. Solving this and substituting back gives you exactly the same 2 solutions like in the video.
I teach courses in data analysis and although my students have studied engineering math up through integral calculus, where they have the most trouble is algebra. I agree that substitution is an effective way to reduce the complication, as well as to save ink, chalk and time. however, I find some students have trouble with this and I think it goes back to the fundamental principles of algebra. Have several students have said they are uncomfortable with expressions that contain both letters and numbers. I appreciate your examples which clearly illustrate why substitution is a good idea, but I’m still having trouble with students that are extremely hesitant to do this.
I'm in awe of all the skill shown here. Personally, I don't like to make my brain hurt, so I just use a solver, like Matlab Symbolic Toolbox (There are others equally as good like Mathematica, Maple, MathCAD, PTC, etc.) Here's the solution in Matlab in 117.8 milliseconds: tic syms x eqn = 2^x - 3^x == sqrt((6^x - 9^x)) S = vpasolve(eqn,x,-3) toc S = -1.709511291351454776976190262174 Elapsed time is 0.117857 seconds
I would say Let u = 2^x - 3^x We find 3^x * u = 6^x - 9^x Rewriting, we get u = sqrt(3^x * u) Equivalently, we get u = 2^x - 3^x = 3^x So, 2^x = 2*3^x, and we solve from there.
I just substituted y=2^x - 3^x so you get y = sqrt(y*3^x)...divide by sqrt(y) on both sides then square both sides. Substitute back in and divide both sides by 2 and you get 2^(x-1)=3^x then you do some rearranging and use change of base for easier reading and the answer is 1/(1-log2(3)). The other solution of x=0 is easy enough to see in the original equation, since both sides are effectively differences of exponentials in x.
substitution seems to be an unneccessary complication. after extracting srqrt(3^x) out of the sqrt you can divide the whole thing with sqrt(2^x - 3^x) and you get sqrt(2^x - 3^x) = srqrt(3^x) take the square and after some manipulation you get (2/3)^x = 2
first method over-complicated things imo. for one, you didnt really need u-substitution, you couldve just factorised a^2-3ab+2b^2=0 and made it (a-2b)(a-b)=0 and derived the two exponential equations from there. additionally, at 6:40, it wouldve been far easier to have simply put (2/3)^x=2 into log form straight away, thus finding x immediately. i.e., you couldve just found x to be log base 2/3 of 2. besides that, good vid, and interesting question!
It can also be solved by changing the base of 2^x to 3^(x•a), just find the value of a (which is log2/log3) and cancel the common terms. A simple quadratic equation is obtained as t-1 = √(t-1) => t = 0 or 1; where t=3^(x(a-1)) -1 Now substitute t and solve to get the same answers!
I did it similarly: I got to when a^2 - 3ab + 2b^2 = 0. But, I did the quadratic formula in terms of a. a = (3b +- sqrt(9b^2 -4(1)(2b^2)))/(2*1) a = (3b +- sqrt(b^2))/2 a = b, 2b Then, I finished the problem like you did in method 1.
My method was: Given: 2^x - 3^x = sqrt(6^x - 9^x) Solution: Squaring both sides and by the rules of exponents we have, 2^2x -- 3^2x = 6^x - 9^x Using natural logs we have, 2x ln 2 - 2x ln 3 = x ln 6 - x ln 9 Cancelling 2x and x we have, ln 2 - ln 3 = ln 6 - ln 9 Using the natural log table we have, 0.6931 - 1.0986 = 1.7918 - 2.1972 Simplifying, -0.4055 ≈ -0.4054 Idk if this is correct or not, it sure was fun though!
I usually prefer log in the base of the most common number,makes it easy to see what I'm working with. log2(2)/( log2(2)-log2(3) = 1/(1-log2(3)) ≈ 1/(1-1.6) ≈ -5/3
Writing a for 2^x and b for 3^x one gets a - b = √(b(a-b)) or √(a- b) = √b or a = 2*b or 2^x =2* 3^x or x = log (2) to the base (2/3) = 1/(1 - log(3) to the base 2) = 1/(1- lg(3) )
@@robertveith6383 Only the x is the exponent (that should be clear when you do this calculation for yourself), hence no grouping symbols are needed here.
@Janina: I first tried squaring both sides, but after solving the equation, I then realized, too, that simply dividing the original equation by 3^x is the fastest method. :)
Sound? I has always been OK for me, except maybe on a rare exception. I'm a bit hard of hearing and in general I hear you (with headphones on) very well with the volume at my default setting. btw, you kept recording 30 seconds after "bye bye".
On the 2nd method I’m confused as to why you square the square root of a-b. It was just a-b before that. Just confused on that step and how it makes it factorable.
6^x - 9^x is obviously negative. The square roots of something negative is not real. This can only happen if 6^x - 9^x is 0 such as 2^x - 3^x. So x = 1 is the only solution
Where is the square root defined....? You just calculate, without asking in which set x is lying. x real, x complex and so on. And you argue: If x is a solution, then x must be... But you have to prove, that this x is actually a solution. For ex: (1) find x,. x*x = - 1, x real; you argue, x^4 = 1 and then x = 1 and x = -1 , if x shall be real. But it is known, that x^2 = -1 hast no real solution. Equation (1) has no real solution... Notice: There are no equivalent operations. Thats why you have to put the x, you found , in (1) and then calculate...
Whose gifted? I'm a gifted, I didn't know that when I was students but where do you want to solve it when using a function which is real? A square root doesn't take a negative number, is it a complex number? No x which is real can answer to that.
I apologise in advance for being rude and monstrous but....gifted students my ass. This is a rather easy problem once you are slightly familiar with algebraic tricks. Calling someone gifted when he's not is harmful for that person. When he realises he's no more gifted than the average person he'll become depressed and frustrated. I've seen that with many kids and to a lesser extent classamates during my uni years. .
Who gives a xxxxxxx about it? It’s not relevant in our live no more longer! What as good as will do in my live anyway! Unnecessary, absolutely rubbish, no need to know and just waist of time..... don’t put on those kinds of rubbish to get a like. Because you know how to solve or you copying someone’s knowledge it doesn’t have to be ours problem.
Sir I must confess that this problem wasn't a test at all comparing with the other problems because if you multiply the lhs by 3topowerx／3topowerx the answer will instantly appear
Here is the tweet from Ahmed Salman that has all the problems:
twitter.com/log2022x/status/1535957868502994946
thanks!
I went straight in with a similar method to your #2. As the LHS is a factor of the expression in the RHS I put
u = 2^x - 3^x
And squared both sides to give a quadratic in x and we get that log expression as the other solution.
Before watching the video:
Given:
2↑x - 3↑x = √(6↑x - 9↑x)
To find:
x
Separating RHS into prime factors to match LHS:
2↑x - 3↑x = √(2↑x·3↑x - 3↑2x)
Taking 2↑x as a and 3↑x as b:
a - b = √(ab - b²)
Squaring both sides:
a² - 2ab + b² = ab - b²
a² - 3ab + 2b² = 0
a² - 2ab - ab + 2b² = 0
a(a - 2b) - b(a - 2b) = 0
(a - b)(a - 2b) = 0
from here, a = b, or a = 2b.
a = b → 2↑x = 3↑x
Taking log of both sides:
xlog2 = xlog3
Moving all terms to LHS:
xlog2 - xlog3 = 0
x(log2 - log3) = 0
As log2 cannot be equal to log3, x = 0.
a = 2b → 2↑x = 2·(3↑x)
Taking log on both sides:
xlog2 = log(2·(3↑x)
Using logab = log a + logb:
xlog2 = log2 + xlog3
Taking all x terms in LHS:
xlog2 - xlog3 = log2
x(log2 - log3) = log2
xlog⅔ = log2
x = log2/log⅔.
Testing,
x = 0 trivially works as it reduces both sides to 0.
When x = log2/log⅔, I'm literally not capable of testing.
Alternative solution:
a - b = √(ab - b²)
a - b = (√b)√(a - b)
(a - b) - (√b)√(a - b) = 0
(√(a - b))(√(a - b) - √b) = 0
a = b or √(a - b) = √b → a = 2b
Proceed as above.
Hey I got both solution paths for a change!
My method is similar to your first method, except that I solved for a in terms of b directly after getting the quadratic. From there, I got the correct solutions.
In your opinion is normal that a 16 years old student can't solve this?
@@dani3672 yeah it's normal until and unless you don't prepare for maths Olympiads
Great video!
Might i suggest you use the letter q for substitution? I think it’s a very under-appreciated letter and it deserves some love
Not to mention g, j, l, w...
q looks like 9 so I try to stay away from that
@@SyberMath
Fair enough. The way i write my Qs, i actually loop the tail of it to the right, almost like a cursive q
@@eliasmazhukin2009
j is definitely another one
Your presentations are just fine. The audio is ok, the visuals are ok, and your pace is just fine.
Keep up the good work.
Thanks, will do!
Nice one!
I have become a HUGE fan of your Math and maybe more.. the below comments are for the "more" part (just on a lighter note, hope you don't mind)
1) You should be awarded a Master's degree - Master of Substitutions
2) You have the knack of somehow bringing in your "2b or not 2b", "Let's call it 'y'; now don't ask me 'why'" puns... So maybe a 'pun'dit of Substitutions
3) I waited till the end of this video - felt almost like waiting for the post-credit scene after a Marvel movie
There can be so much humor in mathematics. It's great.😀🤣😀🤣😀
I solved it the same way as your first method. Although you could easily guess x = 0 as one root just from inspection of the equation.
I didn't start with a substitution. Squared both sides first and distributed the LHS. After distribution you get 4, 6 and 9 as the bases. Moved the whole RHS to the left. Which gives you 4^x-3*6^x+2*9^x=0. Now I divided both sides by 4^x (as 4^x can't be zero) and then substituted (3/2)^x as Z. Which gives you a quadratic equation. Solving this and substituting back gives you exactly the same 2 solutions like in the video.
I teach courses in data analysis and although my students have studied engineering math up through integral calculus, where they have the most trouble is algebra. I agree that substitution is an effective way to reduce the complication, as well as to save ink, chalk and time. however, I find some students have trouble with this and I think it goes back to the fundamental principles of algebra. Have several students have said they are uncomfortable with expressions that contain both letters and numbers.
I appreciate your examples which clearly illustrate why substitution is a good idea, but I’m still having trouble with students that are extremely hesitant to do this.
I'm in awe of all the skill shown here. Personally, I don't like to make my brain hurt, so I just use a solver, like Matlab Symbolic Toolbox (There are others equally as good like Mathematica, Maple, MathCAD, PTC, etc.)
Here's the solution in Matlab in 117.8 milliseconds:
tic
syms x
eqn = 2^x - 3^x == sqrt((6^x - 9^x))
S = vpasolve(eqn,x,-3)
toc
S = -1.709511291351454776976190262174
Elapsed time is 0.117857 seconds
Thank you! 😍
I would say
Let u = 2^x - 3^x
We find 3^x * u = 6^x - 9^x
Rewriting, we get
u = sqrt(3^x * u)
Equivalently, we get u = 2^x - 3^x = 3^x
So, 2^x = 2*3^x, and we solve from there.
I just substituted y=2^x - 3^x so you get y = sqrt(y*3^x)...divide by sqrt(y) on both sides then square both sides. Substitute back in and divide both sides by 2 and you get 2^(x-1)=3^x
then you do some rearranging and use change of base for easier reading and the answer is 1/(1-log2(3)). The other solution of x=0 is easy enough to see in the original equation, since both sides are effectively differences of exponentials in x.
substitution seems to be an unneccessary complication.
after extracting srqrt(3^x) out of the sqrt you can divide the whole thing with sqrt(2^x - 3^x) and you get sqrt(2^x - 3^x) = srqrt(3^x)
take the square and after some manipulation you get (2/3)^x = 2
Thanks, much more straightforward
@chaparral82: Even easier: divide the whole equation by 3^x from the start, then you arrive at (2/3)^x - 1 = 0 or = 1 very fast.
first method over-complicated things imo. for one, you didnt really need u-substitution, you couldve just factorised a^2-3ab+2b^2=0 and made it (a-2b)(a-b)=0 and derived the two exponential equations from there. additionally, at 6:40, it wouldve been far easier to have simply put (2/3)^x=2 into log form straight away, thus finding x immediately. i.e., you couldve just found x to be log base 2/3 of 2. besides that, good vid, and interesting question!
What graphing software or whatever, do you use?
desmos.com
It can also be solved by changing the base of 2^x to 3^(x•a), just find the value of a (which is log2/log3) and cancel the common terms. A simple quadratic equation is obtained as t-1 = √(t-1) => t = 0 or 1; where t=3^(x(a-1)) -1
Now substitute t and solve to get the same answers!
I did it similarly:
I got to when a^2 - 3ab + 2b^2 = 0.
But, I did the quadratic formula in terms of a.
a = (3b +- sqrt(9b^2 -4(1)(2b^2)))/(2*1)
a = (3b +- sqrt(b^2))/2
a = b, 2b
Then, I finished the problem like you did in method 1.
same using the quadratic formula is nice because you can easily get a relation between a and b.
My method was:
Given:
2^x - 3^x = sqrt(6^x - 9^x)
Solution:
Squaring both sides and by the rules of exponents we have,
2^2x -- 3^2x = 6^x - 9^x
Using natural logs we have,
2x ln 2 - 2x ln 3 = x ln 6 - x ln 9
Cancelling 2x and x we have,
ln 2 - ln 3 = ln 6 - ln 9
Using the natural log table we have,
0.6931 - 1.0986 = 1.7918 - 2.1972
Simplifying,
-0.4055 ≈ -0.4054
Idk if this is correct or not, it sure was fun though!
I usually prefer log in the base of the most common number,makes it easy to see what I'm working with.
log2(2)/( log2(2)-log2(3) = 1/(1-log2(3)) ≈ 1/(1-1.6) ≈ -5/3
Writing a for 2^x and b for 3^x one gets
a - b = √(b(a-b))
or √(a- b) = √b
or a = 2*b
or 2^x =2* 3^x
or x = log (2) to the base (2/3)
= 1/(1 - log(3) to the base 2)
= 1/(1- lg(3) )
i have divided both parts by 3^x and ended up with y=sqrt(y), where y=(2/3)^x-1. same answer then :)
If x - 1 is an exponent, then you must have grouping symbols around it because of the Order of Operations.
@@robertveith6383 Only the x is the exponent (that should be clear when you do this calculation for yourself), hence no grouping symbols are needed here.
@Janina: I first tried squaring both sides, but after solving the equation, I then realized, too, that simply dividing the original equation by 3^x is the fastest method. :)
Hi, this one was pretty easy.
I think all of YOUR students are gifted :)
I love the visual presentation of coloured inks on a black background. Please what hardware/software did you use to record the video?
Thank you!
I use an iPad, an Apple Pencil and the Notability App.
I record the videos using screen recording.
0 and log3(2)/log3(2)-1.
I used logs in school, before 1968. Since then, I have no taste for them, except on the slide rule. I arrived at (2^×-3^×)=3, and that was that.
Sound? I has always been OK for me, except maybe on a rare exception. I'm a bit hard of hearing and in general I hear you (with headphones on) very well with the volume at my default setting. btw, you kept recording 30 seconds after "bye bye".
I know! 😂
Thanks
wow what a nice problem! thank you for sharing
No problem 👍
@@SyberMath 👍👍👍
..... (a^2--3ab+2b^2) can be easily factorised as a^2--2ab--ab+2b^2, I. e(a--b) (a--2b). So, a=b and 2b.thereafter follows the solution.
(a-b) = sqrt(b)sqrt(a-b), sqrt(a-b) = sqrt(b). Taking sq, a - b = b. a = 2b, a/b = 2. (2/3)^x = 2. Taking the log, done.
On the 2nd method I’m confused as to why you square the square root of a-b. It was just a-b before that. Just confused on that step and how it makes it factorable.
Excellent! Thank you!
If the base are the same number and the power also similar how we gain the answer. Like 2 to power 3 multiply 2 to power 3
I got a very different answer!
(I also made arithmetic mistakes)
i just simply factorise the RHS, set the equation equal to 0 then easily find two solutions x=0 or x=1/(1-log2(3))
2b, 2u 😂😂😂
"I might get a 2u..."
Just like that😊
And yes, Audio is quite perfect!
Awesome! Thanks.
Some people are annoyed by those but I think the majority is ok
2nd method just x skill minds ... genious 🧐 ... thx teacher !!
Thank you sir♥️
Very nice solutions
Thank you
Volume is fine
You can multiple both side by 3^x and proceeded.
very well done
Merci beaucoup
O hi aayega kyunki 6-9 hamesha negative dega lekin root mein negative value nhi ho sakti that's why only one option if x= o
Sir a-b is not equal to under root a-b whole square
I used Log and found that, log2/(log2-log3) = ln2/(ln2-ln3)
Very nice!
Thanks!
Substitution go brrr!!
Just square both sides of the equation, you'll get a very easy quadratic equation.
Sounds good!
My first instinct was to factor ab-b²
Your volume is too be just fine.
Why do you only upload videos about algebra, it would be good if you uploaded geometry too
Volume is good!
Great!
Nice equation!
Glad you think so!
a=b; and A-b=B
6^x - 9^x is obviously negative. The square roots of something negative is not real. This can only happen if 6^x - 9^x is 0 such as 2^x - 3^x. So x = 1 is the only solution
@ yes, of course, I make this error 😁ops
Весёлый у парня английский. Индус, наверное.
Where is the square root defined....?
You just calculate, without asking in which set x is lying. x real, x complex and so on. And you argue: If x is a solution, then x must be...
But you have to prove, that this x is actually a solution.
For ex: (1) find x,. x*x = - 1, x real; you argue, x^4 = 1 and then x = 1 and x = -1 , if x shall be real. But it is known,
that x^2 = -1 hast no real solution.
Equation (1) has no real solution...
Notice: There are no equivalent operations. Thats why you have to put the x, you found , in (1) and then calculate...
Amazing
Thank you!
Second method is the best
Glad you think so!
x = 0 is a solution by inspection
But not the only one.
Lots of extraneous noise at the end of the video-like you forgot to turn off the mic
Sorry
lol i tried to solve it with derivatives but its so hard on that way and i left it
x ={0, - log2/log3/2}
No problems in hearing what you say
"Substitution forever., right?" 😂👌
Yesss!
Hi . This eguat, is pretty difficult.
x=0?
👍
Cool 🍺🍺🍻.
Hi syber!,nice video.. to be or not to you 😅
Thanks 😅
I like your vids, do not speak louder pls, your soft voice was a big plus
Thank you for the feedback! 🥰
Loud enough
Thanks
Whose gifted? I'm a gifted, I didn't know that when I was students but where do you want to solve it when using a function which is real? A square root doesn't take a negative number, is it a complex number? No x which is real can answer to that.
No, I can't
x = 0
In IRAQ
i solved it on my own imma gifted now???
Yessir
I apologise in advance for being rude and monstrous but....gifted students my ass. This is a rather easy problem once you are slightly familiar with algebraic tricks. Calling someone gifted when he's not is harmful for that person. When he realises he's no more gifted than the average person he'll become depressed and frustrated. I've seen that with many kids and to a lesser extent classamates during my uni years. .
Why don't you post your own videos showing the "tricks"? Not everyone knows them.
CASIO fx-CG500 and done
X=0, -1.71.
The second is approximate.
@@robertveith6383 I can answer it in log terms also. But for understanding of everyone a simple answer.
Lost me at the very start
You explain things in a very muddy way.
Do you have any idea just how ridiculous your 2u and 2b comments sound? I think you should listen to yourself on repeat until you realise.
Who gives a xxxxxxx about it? It’s not relevant in our live no more longer! What as good as will do in my live anyway! Unnecessary, absolutely rubbish, no need to know and just waist of time..... don’t put on those kinds of rubbish to get a like. Because you know how to solve or you copying someone’s knowledge it doesn’t have to be ours problem.
Did exactly the same as Don Ensley ...
By squaring
4^x + 9^x - 2*6^x = 6^x - 9^x
4^x + -3* 6^x + 2* 9^x = 0
(2^x)^2 -3 (2^x)(3^x) + 2* (3^x)^2 = 0
lets define z = (2/3)^x
z2 - 3z + 2 = 0
(z-1)(z-2) = 0
z=1 => x=0
z=2 => x = log(2) / log(2/3) = -1.70951129
Sir I must confess that this problem wasn't a test at all comparing with the other problems because if you multiply the lhs by 3topowerx／3topowerx the answer will instantly appear