I think it's not because they're heavy but because of their temperature. We're taking about an ln triangle and liquid nitrogen is not mechanically problematic.
If u differentiate at the right triangle square law thing, removing the powers, then raise from e as powers removing ln , then solve the quad eqn u get 3/2 which is a correct soln Edit: actually there is some thing wrong with this method though i haven't figured out what I accidentally checked for lnx + ln2x = ln3x rather than the square form, so the soln is wrong Not deleting incase someone wishes to help out
@@thexoxob9448it is possible with complex numbers, and you are then squaring it to make it real. The i-1-0 triangle is an example of absurd triangles you can create like this. The math works out, even if the geometry doesn’t.
5:12 the nice little detail about completing the square here is if you do NOT simplify ln(2) - ln(3) to ln(2/3), but add [ln(2) - ln(3)]^2 to [ln(3)]^2 - [ln(2)]^2 on the RHS, the [ln(2)]^2 will CANCEL 10:00 an approximation is x = 3.8549
I started working this out myself until I reached the quadratic in LnX at which point I realized there was a much easier way to find the solution. All I had to do was watch the video and bprp would work it out for me :)
Dear bprp, I have a question, and I'd appreciate it if you solve it in your next video: Find the max/min value for sinA*sinB*sinC where A, B, and C are three angles in a triangle (A+B+C=pi) Thank you
Possible solution?: since A, B and C are angles which form a triangle, you could take C=π-(A+B). Then, sinC= sin(A+B) due to allied angles. Resulting expression is sinA*sinB*sin(A+B). I used maxima and minima for above expression using partial derivatives and got the answer.
I've started watching your vids since a month and the way u explain is so cool. i could understand understand calculus at the age of 14 thanks to you! #yay
Thanks for this vid. I solved something similar inspired by this problem: instead the sides of the triangle were cosh x, cosh(2x) and cosh(3x). Took a lot of algebraic manipulation but the final answer was pretty cool. Maybe another video?
It's simply fascinating how the quadratic formula pops up like this. More than once a non-scientist/engineer/mathematician has said to me, "They made us memorise the quadratic formula in school. Why? Where will that ever be relevant?" And the answer is... well, everywhere! If you could pick only one formula to memorise, I think this would be a strong choice!
This x is actually a solution to log(x)^2 + log(2x)^2 = log(3x)^2 for any log base greater than one. Bases between one and zero satisfy the equasion but they don't make a right triangle as log(x) would be negative. The other solution of the quadratic formula will give the right answer for bases between one and zero.
Note: x needs to be greater than 1.5 as sum of two sides need to be greater than third side or the difference between 2 sides needs to be less than the third side.
At 9:09 you discard the negative sqrt. But this leads to a positive value of x: (3/2) exp( -sqrt( ln(3/2) ln(9) ) ) = .536676; (3/2) exp(+-sqrt( ln(3/2) ln(9) ) ) = 3.854877
Even though it's a positive value of x, the side length of the triangle which is ln x will become negative. And we can't have triangle with negative sides
Has anyone noticed the WIZARDRY at the first 3 seconds of the video?!? I haven't yet watched this but the first few seconds scared the bejezsus outta me. Why did they do that? The editor must have had a chuckle.
This is actually one I was able to solve by myself! Very cool, I had to attempt a variety of different methods before thinking to expand ln²2x into (ln2 + lnx)², but once I did that everything was clear.
I saw your great older video on x^x. Would you consider making a video on plotting x^x (in 3 dimensions) for Real input and complex output? I tried to sketch the full 3d curve, with the x axis being Real and running perpendicular to the complex plane which is used for the output of x^x. So the x axis is the Real input; the y axis is the Real output and the z axis is the imaginary output. So the y and z axes form the complex plane output of the Real x input. So you have a simple exponential-looking 2d curve for positive x, it crosses the real y axis (or has a limit at x=0) at y=1, but the curve then becomes a complex shrinking spiralling "vase" shape for negative x. It's the "smoothness" of the curve as it crosses the y axis and changes from Real 2d to Complex 3d that I can't visualize. Would you consider making a video on this 3d graph and discuss the 3d smoothness of the real-complex transition at x=0 ? i.e. what's the limit of the 3d angle of the complex curve at x=0. AND: on this graph is x^x at x=0, a forbidden indeterminate point or is it equal to 1 ?
I believe I messed up somewhere along my working and don't feel like restarting. But from a number theory perspective, couldn't this be solved through Euclid's formula? Often used only with integers, but it applies to the real numbers too. If a^2 + b^2 = c^2. Where: a = m^2-n^2 = lnx b = 2mn = ln2x c = m^2+n^2 = ln3x We can raise everything to the power of e. Then rearrange for x in each equation. And set 3x to be equal to the sum of each equation. (3x = e^a + e^b + e^c). I'm not sure where to go from here though, but I haven't worked through far enough to think about that section, and I'm too lazy to do it since I already mucked up once.
In my take, I factored out the 4 from the square root resulting in the product of the square root and 2 then I factored out 2 in the numerator and cancelled it with the 2 in the denominator. When you didn't do the same I expected you would have some twist so I was afraid if I have to rewrite it again.
If you solve the generalized problem of using sides ln(nx), ln((n+1)x), and ln((n+2)x), you get: x = (n+2)/(n (n+1)) * e^sqrt(2 * ln((n+2)/n) * ln((n+2)/(n+1)))
@@Smosh7i That does work algebraically, but if x < 1, then ln(x) < 0. Geometrically, it doesn't make sense for the edge of a triangle to have a negative length.
I got the same answer, I wanted to check it before watching this video. Checking it with algebra by putting the solution back into the original equation proved difficult, much harder than the actual problem in fact.
Wow!! I solved it by a different method (but your method is much simpler and shorter) and got this answer x = 3^[a(a+ sqrroot2)] Where a = sqrroot {[log₃(3/2)]} I thought my answer is wrong ,but after using the calculator, I found that my answer is correct !! 🥳🥳
I knew I was wrong when the answer I got was super long, roughly 3 times longer than the one you got. I checked both of their exact values to make sure it wasn't just a different way of expressing the same value and it wasn't :(
Bprp's top 10 catchphrases 1. Let's do some math for fun! 2. Oh my god! Looks pretty crazy! 3. Wouldn't it be nice... 4. Don't forget the plus C! 5. Today, we have the integral of... 6. Let's go to the complex world! 7. I don't like to be at the bottom, I like to be on the top. 8. Bring this down down! 9. Don't worry, don't worry. 10. The best friend of the black pen is the red pen.
@@shivamchouhan5077 in some countries (like Poland, where I live, for example) people use commas to mark the decimal point and use dots in big numbers e.g. 1.000.000
Ln9 can be 2ln3 so u simplify 2 and 1/2 and you'll get (ln3)² and you simplify the square root and ull get e^(ln3) and you simplify more and you'll finally get 9/2
Great stuff here. When I saw the thumbnail my first thought was IS THIS POSSIBLE? Any other type of functions we can use for the sides of the right angled triangle? What about e^x?
ln(3x)^2 = ln(2x)^2 + ln(x)^2 Then take the derivative of both sides above and cancel common factors ln(3x) = ln(x) + ln(2x) ln(3x) = ln(2x^2) ln(3x/2x^2) = 0 ln(3/(2x)) = 0 then raise to the power of e 3/(2x) = 1 x = 3/2
Applying derivatives to both sides of the equation does not produce an equivalent equation. For example equation x = x + 1 doesnt have solutions, but if you take derivatives it will become 1 = 1, i.e. it is correct for all x.
The other issue here (besides the one Andrey pointed out-- I think you meant square root, not derivative) is that you still have to take (ln3 + lnx)^2 = ln3ln3 + 2ln3lnx + lnxlnx. (p + q)^2 =/= p^2 + q^2 sqrt(p^2 + q^2) =/= p + q (p + q)^2 = p^2 + 2pq + q^2 By Pythagoras (substituting lnx = z, ln3 = a, ln2 = b for the sake of everyone's sanity) we expand our brackets and reach: z^2 + 2az + a^2 = 2z^2 + 2bz + b^2 Gathering, simplifying, and substituting back a and b: z^2 + ln(4/9)z + ln(6)ln(2/3) = 0 As you can see, it's a quadratic with two solutions and no common factors to cancel. If you took the exponential of both sides now you'd reach an impasse (or at least something very gross). Remember you have to take exp of the entire expression, and by log rules, those multiplying lns would end up as powers: alog(b) = log(b^a) lnxlnx = ln( x^(lnx) ) 1 = exp{ ln[ x^(ln(x)) × x^ln(4/9) × ... Grim! 😅
Now I'm wondering whether there are any "log-Pythagorean triples", i.e. integers a, b, c such that (ln a)^2 + (ln b)^2 = (ln c)^2. If there are, how would one go about finding them?
I tried this just now; first of all, thought it was impossible - then noticed my basic algebra error! I got x = 0.2406 (4 decimal places). Similar method; use Pythagoras then simplify to get quadratic in exp(2x) giving exp(2x) = (1 + sqrt(5))/2 (I think this is correct)
This is not too hard. Just apply the Ln formula, solve a quadratic equation, and take the exp. A year 11 should be able to do it. Takes less than a page. Great exercise and great content.
Almost the same as BPRP direct analysis, notice that: log(2x) - log(x) = log(2) log(2x) + log(x) = log(2x^2) from which the product gives the difference of squares, [log(2x)]^2 - [log(x)]^2 = log(2)log(2x^2) = log(2)[log(2)+2log(x)] ...eq(1) from the triangle pythagoras, [log(2x)]^2 + [log(x)]^2 = [log(3x)]^2 = [log(x) + log(3)]^2 ...eq(2) eq(2) - eq(1) gives, 2[log(x)]^2 = [log(x) + log(3)]^2 - log(2)[log(2)+2log(x)] a quadratic in log(x), let u = log(x), u^2 - 2[log(3/2)]u - log(6)log(3/2) = 0 solve for u using quadratic formula and your done x = e^(1/2){ 2log(3/2)+- sqrt[4[log(3/2)]^2 +4log(6)log(3/2)] } etc..
Excellent video, but I will say that I feel like one of the solutions is missing. While I understand why you made the decision to make ln(x) strictly positive, I feel like it's more in the spirit of math to consider the negative solution as well. When I did it I interpreted a negative length to be a normal length but scaled in the opposite direction, and thus drew the triangle upside down. When you draw it out, it's a totally valid right triangle.
I like crazy pythagorean triple questions. I have one for you... Can you find a pythagorean triple (a, b, c) such that (1/c, 1/b, 1/a) is also a pythagorean triple?
Just a tip for quadratic equations: use simplified form of the solutions when coefficient of the linear term is even as it was the case here, i.e,: ax + 2bx + c =0 => x= [-b +|- sqrt(b^2 - ac)]/a 😉 With a=1, even simpler x= - b +|- sqrt(b^2 - c)
@@NoNameAtAll2 no I didn’t. That’s the whole point. It is simplified. You don’t have the division by 2. The coefficient of at x is even: 2b, thus -2b/2a= -b/a and sqrt[(2b)^2 - 4ac]/2a= sqrt[b^2 - ac]/a With the coefficient a of x^2 being a=1 you have the simplified solution as indicated in the comment above ☝️
Neat problem - more subtle than I at first thought - which was that you would be proving an identity. The solution can be summarized as follows : put a(x) = ln( x ) ; b(x) = ln( 2*x ) ; c(x) = ln( 3*x ) ; if a(x)^2 + b(x)^2 = c(x)^2 -> x = { 3*N/2, 3/(2*N) } where N = e^y ; y = sqrt(-2*ln(2)*ln(3)+2*ln(3)^2) -> N = 2.569917715.. x = { 0.5836762755, 3.854876572 } The open question is : when are mathematicians going to admit that not only are calculators here to stay but also math software ?
I love your videos so much! I wish as a young nerd interested in math I had such wonderful resources available to me. Unfortunately I was a young nerd is 80's America, before the internet was common let alone RUclips and pretty much the worst time to be young nerd interested in math. :-/
Because the equation he is trying to solve is not a quadratic equation in terms of x, so x is not equal to the solution of the quadratic equation. What he actually did is replace ln(x) = y, then solve the quadratic equation for y, then replace it back.
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I'll ask again: Are you okay?
Thanks for the video
How to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?
Is brilliant worh
Brilliant
Log triangles are naturally very hard to manipulate, on account of their large size and weight.
I agree
I never wood've thought about that.
I think it's not because they're heavy but because of their temperature. We're taking about an ln triangle and liquid nitrogen is not mechanically problematic.
Log triangles are good for the environment. They trap a lot of CO2 in them.
"naturally" But natural logs are a bit easier to work with than other logs.
Fun fact: If you try the same thing with sine you will get x=pi/6 and with cosine x=pi/4
How?
Probs u use trig. Identities
For a sin(x), sin(2x) and sin(3x) triangle and then you got the cos(x) solution as a result of the trig identity?
If u differentiate at the right triangle square law thing, removing the powers, then raise from e as powers removing ln , then solve the quad eqn u get 3/2 which is a correct soln
Edit: actually there is some thing wrong with this method though i haven't figured out what
I accidentally checked for lnx + ln2x = ln3x rather than the square form, so the soln is wrong
Not deleting incase someone wishes to help out
@@astha_yadav who asked🤣he asks about the trig version
You should change the side length to ln(3x), ln(4x) and ln(5x) to make people to remind of the famous 3-4-5 right angled triangle. :)
WolframAlpha gives two solutions for log^2(3 x) + log^2(4 x) = log^2(5 x)
x≈0.25848
x≈0.67166
The 0.25 solution doesn't work because 3 times that is less than 1.. which means length is negative, which is impossible
@@thexoxob9448it is possible with complex numbers, and you are then squaring it to make it real. The i-1-0 triangle is an example of absurd triangles you can create like this. The math works out, even if the geometry doesn’t.
@@thexoxob9448less than 1 isn't negative. it's between 0 and 1
The "Fading In" Intro is so much better!
Thanks!!!
5:12 the nice little detail about completing the square here is if you do NOT simplify ln(2) - ln(3) to ln(2/3), but add [ln(2) - ln(3)]^2 to [ln(3)]^2 - [ln(2)]^2 on the RHS, the [ln(2)]^2 will CANCEL
10:00 an approximation is x = 3.8549
@CaradhrasAiguo49 I agree, that is the number I got! 👍
I did that but still got about 2.45
I agree with your result
I finally managed to get to the solution of the problem all by my self I feel so proud, it is all thanks to your videos
I only feel relieved when I solve maths problems.
BPRP, you are my favorite math teacher. Thanks for another video.
Thank u!
great video! I have been watching you since 2018 and your content is constantly getting better! good job mr. bprp.
Είμαστε συνονόματοι και έχουμε την ίδια εικόνα προφίλ :)
@@athenaP24 don't tell me this is your alt account
Thank you!
@@junuh11 no hahaha
I started working this out myself until I reached the quadratic in LnX at which point I realized there was a much easier way to find the solution. All I had to do was watch the video and bprp would work it out for me :)
0:03 I heard love triangle ...
Dear bprp, I have a question, and I'd appreciate it if you solve it in your next video:
Find the max/min value for sinA*sinB*sinC where A, B, and C are three angles in a triangle (A+B+C=pi) Thank you
Picks complex A, B, C
Possible solution?:
since A, B and C are angles which form a triangle, you could take C=π-(A+B). Then, sinC= sin(A+B) due to allied angles.
Resulting expression is sinA*sinB*sin(A+B).
I used maxima and minima for above expression using partial derivatives and got the answer.
Help me out here. A + B + C = 180° so 180° = pi ??? Am I missing something?
@@davidp4427 radians since we're adults now
0:00 Just a man coming out of the blue with a Blue pen and Red pen and a sweet Log problem for us .
The sudden chimpmunk voice jump scared me at 8:45
0:00 Bro came from imaginary world to real world
😂😂😂
I've started watching your vids since a month and the way u explain is so cool. i could understand understand calculus at the age of 14 thanks to you! #yay
Glad to hear 😃
I used to hate math, but this guy has somehow an interesting way of explaining things, so I somehow just got hooked lol 😂
same, he’s the reason I’m obsessed with math too
Thanks for this vid. I solved something similar inspired by this problem: instead the sides of the triangle were cosh x, cosh(2x) and cosh(3x). Took a lot of algebraic manipulation but the final answer was pretty cool. Maybe another video?
There are 2 real solutions for x
It's simply fascinating how the quadratic formula pops up like this. More than once a non-scientist/engineer/mathematician has said to me, "They made us memorise the quadratic formula in school. Why? Where will that ever be relevant?"
And the answer is... well, everywhere! If you could pick only one formula to memorise, I think this would be a strong choice!
It's even better to understand how to derive this formula! It's pretty easy!
@@Someniatko Yeah, is easy
9:27 you can actually keep the negative since when you exponentiate it will be positive
This x is actually a solution to
log(x)^2 + log(2x)^2 = log(3x)^2
for any log base greater than one.
Bases between one and zero satisfy the equasion but they don't make a right triangle as log(x) would be negative.
The other solution of the quadratic formula will give the right answer for bases between one and zero.
I love these videos
Note: x needs to be greater than 1.5 as sum of two sides need to be greater than third side or the difference between 2 sides needs to be less than the third side.
At 9:09 you discard the negative sqrt. But this leads to a positive value of x: (3/2) exp( -sqrt( ln(3/2) ln(9) ) ) = .536676; (3/2) exp(+-sqrt( ln(3/2) ln(9) ) ) = 3.854877
Even though it's a positive value of x, the side length of the triangle which is ln x will become negative. And we can't have triangle with negative sides
@@shadowgamer6383exactly
Has anyone noticed the WIZARDRY at the first 3 seconds of the video?!?
I haven't yet watched this but the first few seconds scared the bejezsus outta me. Why did they do that? The editor must have had a chuckle.
This was a really good log rule refresher lol
I love how you just pop into existence in the beginning
This is actually one I was able to solve by myself! Very cool, I had to attempt a variety of different methods before thinking to expand ln²2x into (ln2 + lnx)², but once I did that everything was clear.
Amazing creativity
This video was soo satisfying, because I always realised what he was about to do, split seconds before he actually did it
Please please please please please do another marathon session. Really need it. Calculus. Maybe Laplace, Fourier, Bessel etc. Please?
I took me a while to notice how seamlessly he was switching between red and black and now I’m extremely jealous
You forgot to distribute the square power in the b^2 of the cuadratic formula. (2ln(2/3)^2 is 4(ln(2/3))^2, not 4ln(2/3) as you say in the video
Thanks
7:50
Best part: SHWOO!
I saw your great older video on x^x. Would you consider making a video on plotting x^x (in 3 dimensions) for Real input and complex output?
I tried to sketch the full 3d curve, with the x axis being Real and running perpendicular to the complex plane which is used for the output of x^x.
So the x axis is the Real input; the y axis is the Real output and the z axis is the imaginary output. So the y and z axes form the complex plane output of the Real x input.
So you have a simple exponential-looking 2d curve for positive x, it crosses the real y axis (or has a limit at x=0) at y=1, but the curve then becomes a complex shrinking spiralling "vase" shape for negative x. It's the "smoothness" of the curve as it crosses the y axis and changes from Real 2d to Complex 3d that I can't visualize.
Would you consider making a video on this 3d graph and discuss the 3d smoothness of the real-complex transition at x=0 ? i.e. what's the limit of the 3d angle of the complex curve at x=0. AND: on this graph is x^x at x=0, a forbidden indeterminate point or is it equal to 1 ?
cant I use the power rule for logs at 2(ln2)(lnx) so that 2lnx^(ln2) => lnx^(2ln2) => lnx^(ln4)?
That 2 goes to power of not power of lnx so it wiil be (ln(x²))^(ln2)
this is a beautiful problem, your presentation is so impeccable I have watched it several times🙋🏻♂️
Wow, wonderful topic and excellent presentation!
I believe I messed up somewhere along my working and don't feel like restarting. But from a number theory perspective, couldn't this be solved through Euclid's formula? Often used only with integers, but it applies to the real numbers too.
If a^2 + b^2 = c^2.
Where:
a = m^2-n^2 = lnx
b = 2mn = ln2x
c = m^2+n^2 = ln3x
We can raise everything to the power of e. Then rearrange for x in each equation. And set 3x to be equal to the sum of each equation. (3x = e^a + e^b + e^c).
I'm not sure where to go from here though, but I haven't worked through far enough to think about that section, and I'm too lazy to do it since I already mucked up once.
Your videos are amazing!!
In my take, I factored out the 4 from the square root resulting in the product of the square root and 2 then I factored out 2 in the numerator and cancelled it with the 2 in the denominator. When you didn't do the same I expected you would have some twist so I was afraid if I have to rewrite it again.
If you solve the generalized problem of using sides ln(nx), ln((n+1)x), and ln((n+2)x), you get:
x = (n+2)/(n (n+1)) * e^sqrt(2 * ln((n+2)/n) * ln((n+2)/(n+1)))
Tbh in calc 2 it wasnt the calc that got me but the occaisonal algebra trick
The approximate value of x is 3.85488
What about x = 0.583676
@@Smosh7i That does work algebraically, but if x < 1, then ln(x) < 0. Geometrically, it doesn't make sense for the edge of a triangle to have a negative length.
I got the same answer, I wanted to check it before watching this video. Checking it with algebra by putting the solution back into the original equation proved difficult, much harder than the actual problem in fact.
Wow!!
I solved it by a different method (but your method is much simpler and shorter) and got this answer
x = 3^[a(a+ sqrroot2)]
Where a = sqrroot {[log₃(3/2)]}
I thought my answer is wrong ,but after using the calculator, I found that my answer is correct !!
🥳🥳
0:00 man just phased into existence to teach me math 😭
I knew I was wrong when the answer I got was super long, roughly 3 times longer than the one you got. I checked both of their exact values to make sure it wasn't just a different way of expressing the same value and it wasn't :(
You are brilliant👍 ☺ I love your dedication
Bprp's top 10 catchphrases
1. Let's do some math for fun!
2. Oh my god! Looks pretty crazy!
3. Wouldn't it be nice...
4. Don't forget the plus C!
5. Today, we have the integral of...
6. Let's go to the complex world!
7. I don't like to be at the bottom, I like to be on the top.
8. Bring this down down!
9. Don't worry, don't worry.
10. The best friend of the black pen is the red pen.
blackpenredpen always comes up with interesting problems.
Can you explain some math famous problems like the zeta function or something like that
I like this idea
Thank you
After the second step just replace lnx by u and then continue that would be easier
X is approximately 3,8549 and is a transcendental number!
Actually it is 3.854765
@@shivamchouhan5077 Yeah I just rounded it mate
@@desiaasm But you added comma (,) instead of dot(.) So your answer was 38549
@@shivamchouhan5077 in some countries (like Poland, where I live, for example) people use commas to mark the decimal point and use dots in big numbers e.g. 1.000.000
@@nuclear3011 Oh thanks for telling I didn't know that one, btw this can lead to calculations errors in some cases.
Ln9 can be 2ln3 so u simplify 2 and 1/2 and you'll get (ln3)² and you simplify the square root and ull get e^(ln3) and you simplify more and you'll finally get 9/2
Great stuff here. When I saw the thumbnail my first thought was IS THIS POSSIBLE? Any other type of functions we can use for the sides of the right angled triangle? What about e^x?
He did e^x in another video.
ln(3x)^2 = ln(2x)^2 + ln(x)^2
Then take the derivative of both sides above and cancel common factors
ln(3x) = ln(x) + ln(2x)
ln(3x) = ln(2x^2)
ln(3x/2x^2) = 0
ln(3/(2x)) = 0 then raise to the power of e
3/(2x) = 1
x = 3/2
Applying derivatives to both sides of the equation does not produce an equivalent equation. For example equation x = x + 1 doesnt have solutions, but if you take derivatives it will become 1 = 1, i.e. it is correct for all x.
The other issue here (besides the one Andrey pointed out-- I think you meant square root, not derivative) is that you still have to take (ln3 + lnx)^2 = ln3ln3 + 2ln3lnx + lnxlnx.
(p + q)^2 =/= p^2 + q^2
sqrt(p^2 + q^2) =/= p + q
(p + q)^2 = p^2 + 2pq + q^2
By Pythagoras (substituting lnx = z, ln3 = a, ln2 = b for the sake of everyone's sanity) we expand our brackets and reach:
z^2 + 2az + a^2 = 2z^2 + 2bz + b^2
Gathering, simplifying, and substituting back a and b:
z^2 + ln(4/9)z + ln(6)ln(2/3) = 0
As you can see, it's a quadratic with two solutions and no common factors to cancel.
If you took the exponential of both sides now you'd reach an impasse (or at least something very gross). Remember you have to take exp of the entire expression, and by log rules, those multiplying lns would end up as powers:
alog(b) = log(b^a)
lnxlnx = ln( x^(lnx) )
1 = exp{ ln[ x^(ln(x)) × x^ln(4/9) × ...
Grim! 😅
Now I'm wondering whether there are any "log-Pythagorean triples", i.e. integers a, b, c such that (ln a)^2 + (ln b)^2 = (ln c)^2. If there are, how would one go about finding them?
Bonus points if you use Lambert's W function
0:00 It’s true, bprp has super speed.
What about exponential triangle problem, exp(x), exp(2x), exp(3x) ?
I tried this just now; first of all, thought it was impossible - then noticed my basic algebra error! I got x = 0.2406 (4 decimal places). Similar method; use Pythagoras then simplify to get quadratic in exp(2x) giving exp(2x) = (1 + sqrt(5))/2 (I think this is correct)
That was actually a pretty fun problem.
Did he forget the + between the 3/2 and e at the final answer or did I miss sth?
This is not too hard. Just apply the Ln formula, solve a quadratic equation, and take the exp. A year 11 should be able to do it. Takes less than a page.
Great exercise and great content.
Almost the same as BPRP direct analysis,
notice that:
log(2x) - log(x) = log(2)
log(2x) + log(x) = log(2x^2)
from which the product gives the difference of squares,
[log(2x)]^2 - [log(x)]^2 = log(2)log(2x^2) = log(2)[log(2)+2log(x)] ...eq(1)
from the triangle pythagoras,
[log(2x)]^2 + [log(x)]^2 = [log(3x)]^2 = [log(x) + log(3)]^2 ...eq(2)
eq(2) - eq(1) gives,
2[log(x)]^2 = [log(x) + log(3)]^2 - log(2)[log(2)+2log(x)]
a quadratic in log(x), let u = log(x),
u^2 - 2[log(3/2)]u - log(6)log(3/2) = 0
solve for u using quadratic formula and your done
x = e^(1/2){ 2log(3/2)+- sqrt[4[log(3/2)]^2 +4log(6)log(3/2)] } etc..
I thought he said, "Love triangle" 😂🤣🤣
Shouldn’t the final solution be
x = (3/2)+e^(sqrt(ln(3/2)*ln9)) ?
Why don't you set y-lnx from the beginning?
What song is used during the Brilliant ad?
what a incredible video..
Excellent video, but I will say that I feel like one of the solutions is missing. While I understand why you made the decision to make ln(x) strictly positive, I feel like it's more in the spirit of math to consider the negative solution as well. When I did it I interpreted a negative length to be a normal length but scaled in the opposite direction, and thus drew the triangle upside down. When you draw it out, it's a totally valid right triangle.
How to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?
I like crazy pythagorean triple questions. I have one for you...
Can you find a pythagorean triple (a, b, c) such that (1/c, 1/b, 1/a) is also a pythagorean triple?
The golden ratio rules!
What is the answer plz
.
Just a tip for quadratic equations: use simplified form of the solutions when coefficient of the linear term is even as it was the case here, i.e,:
ax + 2bx + c =0
=> x= [-b +|- sqrt(b^2 - ac)]/a 😉
With a=1, even simpler
x= - b +|- sqrt(b^2 - c)
Much more preferable to divide by the common factor than memorizing yet another formula
i know it as x²+px+q => -p/2±sqrt((p/2)²-q)
@@anastasissfyrides2919 it’s not memorising new formula, it’s simplifying the 2’s
- b/2
you forgot to divide b by 2
@@NoNameAtAll2 no I didn’t. That’s the whole point. It is simplified. You don’t have the division by 2.
The coefficient of at x is even: 2b, thus -2b/2a= -b/a and
sqrt[(2b)^2 - 4ac]/2a= sqrt[b^2 - ac]/a
With the coefficient a of x^2 being a=1 you have the simplified solution as indicated in the comment above ☝️
Let x= e^u for an easier time... Great video though :D
Neat problem - more subtle than I at first thought - which was that you would be proving an identity.
The solution can be summarized as follows :
put a(x) = ln( x ) ; b(x) = ln( 2*x ) ; c(x) = ln( 3*x ) ;
if a(x)^2 + b(x)^2 = c(x)^2 -> x = { 3*N/2, 3/(2*N) } where N = e^y ; y = sqrt(-2*ln(2)*ln(3)+2*ln(3)^2) -> N = 2.569917715..
x = { 0.5836762755, 3.854876572 }
The open question is : when are mathematicians going to admit that not only are calculators here to stay but also math software ?
Can you do 100 Linear Algebra video
Well educative
At 7:01 isn’t wrong? Because 2/3 /6 is equals 4. It’s same as 2/3 / 1/6 so we flip second and change the operation into multiplication
0.6666.../6 = 0.1111... not 4
@@Noctarc oh yes i get it now, thank you. I wrote it as 2/3.1/6 but forget that it is 2/3//6-double lines is for longe division line-.
Challenge: find x such that: log(ax)^2 + log(bx)^2=log(cx)^2 for arbitrary a,b,c
x = panda
Hypotenuse and legs are on the both side of the triangle.
Pretty cool algebra 2 problem.
you should simplify the exponential of the square root.
Help me to find integral of x^2/x^4+3 im in distress
For everyone,, the answer is x = 3.8548
btw, thats the form of primes as logarithms of base prime exponents
but the primes are just exponent one, of one ln base
Your t-shirt made me think that the golden ratio will appear in the answer...
couldnt you just exponentiate to get rid of the ln? e^ln(x)^2+e^(ln(2x)^2=e^ln(3x)^2??
dang the intro is smooth
Thanks!
I love your videos so much! I wish as a young nerd interested in math I had such wonderful resources available to me. Unfortunately I was a young nerd is 80's America, before the internet was common let alone RUclips and pretty much the worst time to be young nerd interested in math. :-/
Can someone explain to why why instead of x is equal to in the quadratic formula, he did ln(x) is equal to?
Because the equation he is trying to solve is not a quadratic equation in terms of x, so x is not equal to the solution of the quadratic equation. What he actually did is replace ln(x) = y, then solve the quadratic equation for y, then replace it back.
whats the answer
Guys this solution works! My love triangle problem is gone, thanks to this.
nice job
Yea, nice problem🥳🤔
After a very exhaustive effort, I got the following solution:
x=e^{-ln(2/3) (+/-) sqrt[2ln(3/2)ln3]}
Hi thanks for the vid have a nice day it was great
Gracias.👍🏽
I was really hoping for it to work for all X