the famous equation b^x=log_b(x)
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We will find b so that the exponential function y=b^x is tangent to the logarithmic function y=log_b(x). The key to this problem is to know how to solve equations in the form of f(x)=f^-1(x), namely the original function is equal to its inverse.
Related videos
move e^x horizontally so it touches ln(x): 👉 • e^x meets ln(x)
solving e^x=ln(x) complex solutions: 👉 • Is e^x=ln(x) solvable?
Check out @MuPrimeMath's video on solving f(x)=f^-1(x) • When does f(x)=f⁻¹(x) ...
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0:00 check out KiwiCo
1:00 want b so that b^x is tangent to log_b(x)
4:36 the condition we need
#calculus #hardmathproblem
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wow nice
Hi, I have a question for you: in America if you want to study mathematics can you choose to do or not calculus 3 or it is mandatory?
Fun fact: exp(1/e) is also the largest real number for which the infinite power tower x^x^x^... converges. The smallest is exp(-e). This was proven (iirc) by Euler around the 18th century, and is one of my favorite pure-math theorems.
3b1b
It’s also the maximum value of the function f(x) = x^(1/x). :)
@@Deathranger999 Isnt that already the definition of e?)
@@neoxus30 No. There are multiple different definitions for e but "the point at which the function f(x) = x^(1/x) takes its maximum value" is not a definition I've seen. Two common definitions I've seen are sum_{n = 0}^{infinity} 1/(n!), or limit as n goes to infinity of (1 + 1/n)^n.
@@Deathranger999 one uncommon definition of e that I like is "the number such as the integral from 1 to it of 1/x dx is equal to 1", because this definition has a nice visual interpretation, that can be seen here: kzhead.info/sun/d5GxY8l5en-ido0/bejne.html My second favorite is the infinite sum definition, because it's relatively simple to derive/express and it's very general. For instance, one can calculate e to the power of a MATRIX using it. How cool is that? 3b1b has a great video about this: kzhead.info/sun/gpxuf7t6gWSXrKc/bejne.html
If anyone is wondering, for b
A much more simpler solution 🙃 It's obvious and is pretty known by jus looking the graph that both these graphs pass through y=x Consider the curves, y=x and y=aˣ ⇒ x = aˣ Upon differentiating with respect to x ⇒1 = aˣlna Substitute aˣ as x ⇒1 = xlna ⇒1 = lnaˣ (Rule : xlny = lnyˣ) Again Substitute aˣ as x ⇒1 = lnx ⇒ x = e So value of x is e ⇒ e = a^e Taking ln to the base a on both sides ⇒ ln(e) to the base a = e ⇒ ln(e)/ln(a) = e ⇒ ln(a) = 1/e ⇒ a = e^(1/e) PS : Hope y'all enjoyed this version of versatile solution :)
Woah 😮🔥
Nice and compact 🔥
Another really quality video! You inspire me to make my own content! Thank you! ❤
What a coincidence, today I just came up with and solved this problem: for which a are e^(x-a) and ln(x+a) tangent to each other? And at what point do they touch? I believe it's a really nice question, maybe you could solve it in another video.
I actually solved that a while ago. Link in description 😃
@@blackpenredpen Wow cool! Also, thanks for replying😁 I really like your videos, they are very interesting and entertaining
I love this problem! I actually came up with this exact problem idea a few months ago, and it was very rewarding to solve on my own.
This was my approach to the problem The Question: For what [b] value is the function [b^x] tangent to its inverse? Text Notation: . y'=dy/dx . b^x=exp(x*ln(b)) . log_b(x)=ln(x)/ln(b) Given Observations: 1) b^x=log_b(x), for some [x] value 2) (b^x)'=(log_b(x))', for some [b] value Observational Implications (p1): . exponentiate both sides of the equation: b^b^x=x . substitute [b^b^x] in for [x]: b^b^b^b^x=x . repeat forever: b^b^b^b^...=x . substitute [x] in for [b^b^b^...]: b^x=x Observational Implications (p2): . expand both sides of the differential equation: ln(b)*b^x=1/(ln(b)*x) . substitute [x] in for [b^x]: ln(b)*x=1/(ln(b)*x) . multiply both sides of the equation by [ln(b)*x]: (x*ln(b))^2=1 . take the positive square root on both sides of the equation: ln(b)*x=1 . apply the exponential function to both sides of the equation: exp(ln(b)*x)=exp(1) . simplify: b^x=e . substitute [b^b^b^b^...] in for [x]: b^b^b^b^b^...=e . substitute [e] in for [b^b^b^...]: b^e=e . raise both sides of the equation to the [1/e] power: b=e^(1/e) The Solution: . substitute [e^(1/e)] in for [b]: The function [(e^(1/e))^x] is tangent to its inverse, [ln(x)/ln(e^(1/e))] . simplify the equation: Therefore, the function [exp(x/e)] is tangent to its inverse, e*ln(x)
My approach, calculus by brute force: - If f(x) = f-1(x) then f(x) = x therefore we will solve b^x = x - By definition, this problem only has one solution, as the curves only intersect in one point. - If we try different values of b, we find that 1.44 < b < 1.45 - Since it's about logarithm and exponentials, it's likely that the solution has something to do with e - Try different expressions that contain e - Plug in b = exp(1/e), it works. - There is only one solution, therefore, since exp(1/e) satisfies b^x = x for only one value of x, it is the only solution.
I came up with this question and discussed with my friend about 3 months ago, it took us an entire day to figure out b^x = x holds true lolol Edit: At that time we still haven't started calculus and all my calculus knowledge are from bprp
How lol ?One look at the graph and it should be obvious
@@goddosyourself7970 we drew it very not in proportion and did not use computer for it
@@vitalsbat2310 yeah but isnt it logical evwn if not drawn correctly?I personally thought it was
What school are you attending?
@@aronbucca6777 lol at this current moment I have not learnt any calculus from school and all my calculus is only from bprp, I start calculus the upcoming year I think
Hi BPRP! You should make a video on solving definite integrals using the limit definition. I don't know if this is something you teach but I'm personally having an issue with it in my equivalent Calc 2 course, specifically when I need to solve trigonometric definite integrals. For example, you could make a video on: the integral from 0 to pi/2 of cosxdx
7:09, sooooooooo funny! I’ve had moments like that too. Haven’t we all? I’m glad you left that bit in and didn’t edit it out.
😆 glad u enjoyed it
At 2:52 I don’t know why but I laughed so much 😂😂
I came up with the solution at 1:50 but only because I knew the value of e^(1/e) That was a good intuition! ^^
From knowing values, I first guessed the solution to be √2 and wasted quite some time testing this supposed solution.
@@peterjansen7929 It could have been pi^(1/pi) too. But with regard to the studied functions e^(1/e) was way more probable!
@@marchenwald4666 Undoubtedly! But I didn't know the value of that.
Interesting question!! Now I wonder other relations and inverses we could make tangent to each other... y=(x-a)^2 and (y-a)^2=x? y^a=bsinx and bsiny=x^a? Or any other relation...? I wonder if a method could be generalised for any pair of inverse relations...
by symmetry, notice that at every point where b^x = ln(x)/ln(b), that x=y. for this reason, we can forget the second function and just say: b^x = x. for which b is this x unique? lambert W i suppose comes to the rescue, but yeah
Could you do a problem involving the Schrodinger equation? It could be interesting to see you solve the 2nd order PDE
I d'love to see it too
The answer also comes from the infinite tetration where 1/e^e≤x≤e^(1/e). If you just plug e^(1/e) to b it works
TETRATION? I never realised tetration had an actual use! How on earth do you calculate an infinite tetration?
@@user-mq3um5iu2q kzhead.info/sun/ot2Ym7Sxi6WqpKM/bejne.html This is a video he maid a year ago about the infinite tetration or the infinite power tower
b^x=log_b(x) since they're inverse functions they'll meet at y=x so b^x=x Then x=-W(-ln(b))/ln(b) but for it to exists the argument (-ln(b)) >= -1/e and so b
i solved in a very similar way, thos function are inverse of each others, so the derivate of both side must be equal to one, ln(b)*b^x=1 also the value of b^x is also log_b(x) and with log rules is the same as ln(x)/ln(b) (if b is not e). so i have: ln(b)*b^x=1, b^x=ln(x)/ln(b) i replace the b^x and i have ln(b)*ln(x)/ln(b)=1 so ln(x)=1 so x=e. after that is exacly the same to solve for b
دەست خۆش thank U
I have a question for you: can you solve for b such that log base b of x and b^x intersect twice, with Δx of the interceptions = e
we can find 2 solution by using another methode ! (there might be more solution if my verification was wrong) ln(b)*b^x = ln(x) x = e^(ln(b)* b^x ) = b^b^x so we replace x => x = b^b^b^b^... whe know that if the sol x is unique with a fixed b we have: b^x - logb(x) = 0 and (b^x - logb(x)) ' = 0 so we replace x again => (b^x - logb(x)) ' = ln(b)*b^x - 1/( ln(b)*x) = 0 => ln(x) = 1/ln(x) so x in {e^-1, e^1} Then we take the recurcive sequence u(0) = b and u(n+1) = g( u(n) ) with g(x) = b^x we solve lim u(n) = e^-1=g(g(e^-1)) => b=e^-e (beacuse e^-1 g(x) decrease) and lim u(n) = e=g(e) => b=e^(1/e) (beacuse e^(1/e)>1 => g(x) increase) so the two sol are: b=e^-e at x=e^-1 and b=e^(1/e) at x=e
thanks, so nice ))) could we have some videos on nonintegers in weird places? like e^e^e power tower which has e height
I did it like this: b^x=log_b(x) => b^b^x=x [1] Derivative =1 => b^x*ln(b)=1 => ln(b) = 1/b^x => e^(1/b^x) = b => b^b^x = e [2] [1] and [2] => x=e b^x=x becomes b^e=e Take ln on both sides and you get b=e^(1/e)
with X = exp(w(1/ln b)) ; X = e and b = exp(1/e) I could make that W(e)=1. I didn't know (or didn't remember) it... Thanks!
There should be a "Peyam WOW" at the end of the video!
@blackpenredpen Hi BPRP, at 3:15 , if we don't substitute b^x = log_b x into second equation and manipulate only the second equation directly to use Lambert W as follows:- (b^x)*lnb = 1/xlnb e^ln(b^x) * lnb = 1/xlnb lnb * e^xln(b) = 1/xlnb xlnb * e^xln(b) = 1/lnb W{xlnb * e^xln(b)} = W{1/lnb} xlnb = W{1/lnb} x = W{1/lnb} / lnb is this the same as x = e^W{1/lnb} ?
"typical calc 1 question" *proceeds to pull out the lambert w function*
😂
Back to the old school Bprp look. I like it
😆
Congratulations on getting a sponsor.
very nice
I think, there's an easier way to solve this problem. You only need to calculate the ecuation of an arbitrary tangent line of logb(x) at x = a. Then you know that, at the intersection point of both functions, the tangent line to said point will be y=x so you only need to solve a simple system of equations (to make the equation of the tangent line previously calculated =x) that will lead you to the solution: a=e b=e^(1/e) Nice!
"somewhere between 1.4 and 1.5" me reflexively: "√2"
This is actually beautiful relation
I was wondering this question for almost 2 years, he finally did it !! But I already had the answer from a comment 😅
If you require only that b>0, there is another solution: b = e^(-e). Since b^x is not an increasing function when 0
wait what I looked on desmos and the graphs intersect
@@josenobi3022 But they have the same tangent line at the intersection point. I guess I'm solving the problem "for what b are the curves tangent", which is related but not equivalent to the problem "for what b do the curves have a unique intersection point".
Hey blackpenredpen! I am a huge fan, will do explain this problem for me please? When is the product of X1 and X2 maximum given that the function is f(x)=(X-X1)(X-X2) (derivative is not allowed but I'd like you to do it as if it is at first). I have a divine understanding of parabolas and I only got closer to the answer but never actually got it. I'd appreciate it you would explain that for me, thank you.
Let f be a continuous real-valued function on R^3. Suppose that for every sphere S of radius 1, the integral of f(x,y,z) over the surface of S equals 0. So is it necessary f(x,y,z) must be identically 0?
it is required that the area above the x axis is the same as the area under the x axis, so the center must be at the x, z plane
You know it's serious when blackpenredpen pulls out the blue pen...
this is nice
What a fun exercise.
without watching yet, mi first thought is, that dur to the fact, that log_b(x) is the inverted function of b^x, their meeting point must be also the equal one, as the tangent point of the funktion g(x) = x. So we can start with b^x = x. And the derivative you have ln(b)*b^x = 1. SO you have a system of two equations to solve
There are two solutions, one for x=e and another for x=1/e Initial equation b^x = log(x)/log(b) Partial x derivative log(b) b^x = 1/(x log(b)) Solve the initial equation for b to get b=e^(pog(x log(x))/x), where pog(x) e^pog(x)=x Something to note is that in addition to pog(x) e^pog(x)=x, pog(x e^x)=x (this is true for the purposes of this problem, but not all cases because pog has multiple branches; we only need to consider the 0th branch) Substitute into the partial derivative equation log(e^(pog(x log(x))/x)) (e^(pog(x log(x))/x))^x = 1/(x log(e^(pog(x log(x))/x))) Use log and exponent properties to simplify pog(x log(x))/x e^(pog(x log(x))) = 1/pog(x log(x)) Use pog properties to further simplify log(x)pog(x log(x)) = 1 Assume a solution of the form e^u u pog(u e^u) = 1 u^2=1 u=+-1 x=e,1/e Substituting this in the solution for b x=e,b=e^(pog(e log(e))/e) x=1/e,b=e^(pog(1/e log(1/e))/(1/e)) Simplify pog(e)=1 and pog(-1/e)=-1 x=e,b=e^(1/e) x=1/e,b=e^(-e)
oh god KZhead auto-captioning saved me. Can I hear you? Yeah~
This is 2 impressive curves and more impressive solution ;)
Watching BPRP do a Kiwi Co. Box while doing real analysis is more fun
3:24, clip
VERY INFORMATION VIDEO!!! TANK YU!! (soryr for bad egluish)
2:47 to 2:57 is me doing literally any mathematics
The funny part is that even he got confused about the result 😂😂😂
pure magic
I reduced the system to the equation xlnx = 1/lnb. Then I found that for b>1 this has 1 solution only xo. Then I solved for b = exp(1/(xo•lnxo)) and I used the equation of the first system b^x = lnx/lnb i substituted b = exp(1/(xo•lnxo)) and the equation induced form the substitution is exp(1/lnx) = ln²x • x . An easy solution us x = e so if xo= e from b =exp(1/(xo•lnxo)) then b = exp(1/e). Done
🎶Sometimes when we touch, the honesty's too much 🎵
The first thing I noticed was b^x = x Alas, couldn't get anywhere with it, because I didn't think of derivates at all
Not surprised that bprp tangents the Lambert W 😎😎😎 Cool vid, man! 👍
😆 thanks.
Where you from?
I'm still confused why we can do the trick with y=x... How do you know it's reflected upon exactly that line? An intuitive response would be nice, because it doesn't seem to click for me
because we get the graph of a reciprocal function by subjecting the function to a reflection with respect to the y=x axis
@@noor-dl6im Is this well known if you're more familiar with calc? Because I wouldn't have known that. I'm curious why this is the case aside from just looking at the graph
@@xinzhoupingThis is a fact you usually learn in junior-level high school algebra, that an inverse has a graph that is reflected around the line y=x Suppose an original function is given as f(x), and we call its inverse g(x). For each point on the line of f(x), it will have the coordinates of (x, f(x)). Since an inverse function undoes the function of x, each of these points will also be equivalent to (g(y), y). The plot of the inverse function will have coordinates (f(y), y), which also can be written as (x, g(x)). The fact that you just swap x and y coordinates, means that it reflects around the line, y=0. Note that this isn't a reciprocal function. Reciprocals are a special kind of inverse operation, for multiplication specifically. Inverses in general, are functions that undo other functions.
Thanks. From Indonesian
Can you try solving a question involving Knuth notation? For example: "y = x↑↑x, find dy/dx"?
wait are those defined for non integer values?
Here I am wanting to find the roots of the distance formula.
No way I solved this with guess and check lmao
Solve please 1/x = e^x
pretty cool
V nice found interesting
Can you make a video using the quartic formula ?? 😬😬😬
Please put the music on in the whole video
Once you make the observation that the curves are tangent to each other along the line y = x the problem almost solves itself. b^x·ln(b) = 1/(x·ln(b)) = 1. => x = 1/ln(b) . => b^x = (e^ln[b])^(1/ln[b]) = e . Therefore, e·ln(b) = 1 . => ln(b) = 1/e . Finally, b = e^ln(b) = e^(1/e) ◼
This is great! Nice vid!
WHYYYYYYYYYYYYYYYYY WHY IS IT ALWAYS E THIS IS RIDICULOUS
As a fellow Taiwanese, I see what you did there with mistaking 1 for e :D
Love From India 🇮🇳🇮🇳🇮🇳
love from taiwan
would this also solve e^x = c lnx having one solution
umm what if i put b under them so it will be like b to the b to the x equals to b to the logb x then it will be like b to the b to the x equal to x and uh idk how to expand it from there.im still in secondary school
Better love story than Twilight
👌👌👌
Does this mean that W(1/ln(e^(1/e))) = 1?
Bro please explain what is a logerthem and why is it most widely used in calculus....
A logarithm is a function that takes in a known base and a known answer of an exponent operation, and returns the unknown exponent. It is the inverse function of an exponential function in the form of y = b^x. If we know b and we know x, we can calculate y just as an exponential function. But if we know y and we know b, but don't know x, then we need a logarithm to calculate it. We'd write this with the notation, x = log_b (y), where the _ indicates a subscript, and this is pronounced "log base b of y". There is a special case of logarithms called log base e, or natural log, indicated as ln(x). It turns out that any log base can be written using ln(x), such that log_b (x) = ln(x)/ln(b). Log of the main input, divided by log of the base. Natural log and exponentials of base e are special, because they have the most elegant calculus. With any other base, you end up accumulating a constant multiplier due to the chain rule, so the pure forms of these functions, have base e. The constant in question, is the natural log of the base. So to do calculus on 2^x, you'll first convert it to e^(ln(2)*x). The ln(2) is a constant that you'll accumulate in the chain rule. The reason why it is so widely used in calculus, is that introductory calculus considers the calculus operations on the 5 kinds of "atom functions", called elementary functions, that we use most commonly in math in general. They are as follows: 1. Power functions, such as x^n, where x is a variable and n is a constant power 2. Exponential functions, such as b^x, x is a variable exponent and b is a constant base 3. Sines and cosines from trigonometry 4. Inverse trigonometry 5. Logarithms, that are the inverses of exponentials We also have combinations of the above, using arithmetic and function composition, that are all of interest for calculus.
A reason why logarithms show up so commonly in calculus, where you least expect, is that a special case of the power rule for integration fails, and natural log saves the day. Consider the differentiation power rule, which is d/dx k*x^n = k*n*x^(n - 1). Drop the exponent by one, and have the original exponent join the coefficient. In reverse for integration, this becomes integral k*x^n dx = k/(n + 1) * x^(n + 1). Raise the exponent by 1, and have the new exponent join the coefficient in the denominator. However, this has a problem when n = -1. The new "exponent" would be zero, and we'd divide the coefficient by zero. This doesn't work. Surprisingly though, natural log saves the day. You can show why this works, by starting with n = -1 + h as the original exponent, and taking the limit as h goes to zero: integral x^(h - 1) dx = 1/h * x^h + C Let the constant equal -1/h: (x^h - 1)/h Take the limit as h goes to zero. We have an indeterminate form, and thus can use L'Hospital's rule: d/dh (x^h - 1) = d/dx [e^(ln(x)*h) - 1] = ln(x)*e^(ln(x)*h) = ln(x)*x^h d/dh h = 1 Compose the limit together: ln(x)*x^h / 1 As h goes to zero, x^h goes to 1, as long as x isn't zero. Thus, the limit becomes: ln(x)
Video Suggestion: f'=f^(-1) differential equation
Sort of. It's called a functional equation.
let's have a minute of silence in memory of solution b=e^(-e), which was not found using this approach [*] I am wondering if it can also be found in other way than solving system of equations f(x,b)=g(x,b), d f/d x = d g/d x ?
Lol, I should have known it would be related to e in some way
Simple. b = sqrt(2sqrt(sqrt(sqrt(sqrt(2))))
a^x=b, x=? in terms of a and b sir can you solve this without using logarithms
I'm a Taiwanese student, I'm 14, I like to watch your video because I like to study math. Can you teach Fourier transform? I want to understand this.
I actually have a few videos on it. You can search it on my channel. : )
Here’s a challenge, try to integrate (Cos2x)**0.5 from pi/2 to 0, it’s literally impossible!
Hello, there's a équation im blocked at, its xe^x+x=k so hard
An interesting question i thought of : assume a sequence such as this :√2, ∛3, ∜4,...... and so on where essentially each term of the series is the nth root of n , prove or disprove that any 3 consecutive terms of this series will form an A.P.Using a calculator I notice that the common difference between consecutive terms is approximately the same , but I have no clue on how as to prove or disprove that 3 consecutive terms of this series will form an A.P or how to prove that you cannot prove or disprove it.
Yeah they seem to be in ap (when I tried with a calculator) but only for smaller numbers for example do for 99,100,101 it doesn't work Really nice observation tho
the fast-forwarded clip the W thing whats that?
W(xe^x)=W(x)e^(W(x))=x
e^(1/e)
That's why e memes are so funny
👍👍👍👍👍
First comment?!! It doesn't matter! It's just matter he sends the first ❤️ ! And it's me!! Thank you Teacher ❤️
e to the root of e
I found exp(-e)
Why 1/e? I mean looks helpful but why?
i thought it was root 2
it's easy
why does f(x) = f-1(x) => f(x)=x only if f is increasing? sounds to me like it should work for decreasing as well
Since y = x is always increasing it would make sense that it doesn't work for decreasing. Also it obviously doesn't work for constant functions.
B^(b^x)=x B=x^(b^(-x)) B^(-x)=logx(b) and Im done.
First :D