A Nonstandard Equation

[@ghstmn7320]

Consider f(x) = 4^x -1/x f is differentiable in R* therefore its continuous there. f'(x) = (4^x) ln4 + 1/(x^2) which is positive for all x in R* Since f'(x) >0 and f is continuous, it means that f is always increasing. We can see that f(1/2) =0. Since f is always increasing, its injective therefore it's only zero for x=1/2. So x=1/2 is the only solution to the equation!

[@yoav613]

Just relax and let syber do the hard work for you😊💯

[@SyberMath]

💯☺️

[@creounity]

at first glance, seems x=1/2

[@asheredude]

Same but that's not how you solve

[@creounity]

@@asheredude well once you try making a graph, it's clear that there's 1 intersection point, and it could be guessed in a few seconds :)

[@asheredude]

@@creounity I wrote "same" as I also tried the same way but answer is not the point here the point is to learn new things

[@mcwulf25]

Easy to rearrange so x ln4 e^(x ln4) = ln2 e^(ln2) W both sides and cancel ln2.

[@RSLT]

what a wonderful video with great points.

[@SyberMath]

Glad you enjoyed it!

[@justabunga1]

Another method to do that is to raise both sides by 1/x, which is (1/x)^(1/x)=4=2^2. Therefore, 1/x=2, so x=1/2.

[@Music--ng8cd]

Well done.

[@mcumer]

The easiest solution is x=1/2.. if we consider the function y=4^x-1/x we can notice that the derivative is always >0 ( an exponential function with the base >1 is obviusly an increasing function, therefore with positive derivative, while the derivative of -1/x is 1/x^2, obviusly positive).. we conclude by observing that y can be zero only for x=1/2, because the function is an increasing function

[@SidneiMV]

xln4 = ln(1/x) (1/x)ln(1/x) = 2ln2 ln(1/x) = W(2ln2) a possible value for W(2ln2) is ln2 so a possible solution is ln(1/x) = ln2 => 1/x = 2 *x = 1/2*

[@paraskumar9850]

W sol.

[@josepherhardt164]

Before even watching: Does this smell like a Lambert's W function thing, or what? Of course, you can just stare at this and see x = 1/2 is a solution.

[@thexavier666]

Almost had a heart attack on seeing you cancel the two ln

[@Mediterranean81]

x log (base 4) 4 = - log (base 4) x x= -log (base 4)x plug x=1/2 1/2=log(base 4) 2 1/2=1/2 so 1/2 is a solution

[@scottleung9587]

I got x=1/2 by guessing and checking.

[@barakathaider6333]

👍

[@honestadministrator]

f(x) = 4 ^ x - 1/x f (1/2) = 0 f ' ( x) =( 4 ^ x) ln (4) + 1/x^2 f ' (1/2) =2 * ln (4) + 4 = (ln(2) + 1) * 4 > 0 Hereby f ( x) = 0 at x = 1/2 and this is only minima of this function

[@mcwulf25]

Got it using W

[@RoyLeBlanc]

Enough of that bloody W function!

[@rakenzarnsworld2]

x = 2^(-1)