This is my first post on math stack exchange!

While solving a problem, I came across an equation: $\log(x) = 2x$.

How do I calculate $x$?

actually I was trying to simplify x=100^x. please tell me any other approach, if possible.

  • $\begingroup$ Welcome to the exchange. normally when someone is having problems with an equation like this it would be a good idea to also include what you have tried. That way we can better assist with your issue. $\endgroup$ – Sentinel135 Apr 1 '17 at 4:28
  • $\begingroup$ First step would be to check if (real, I guess) solutions exist at all. $\endgroup$ – dxiv Apr 1 '17 at 4:30

calculus -- ez

let f(x) = 2x, g(x) = log x

f(0) = 0 g(0) = -infinity

f(1)=2 g(1)=0

f(x) and g(x) have no intersection between $0<x<1$ because f(x) only takes on values between 0 and 2 on (0,1), and g(x) only takes on negative values on (0,1).

the rate of change of f(x) is 2 for x>1 {all real x} the rate of change of g(x) is 1/x for x>1 {x>0}

because 2 > 1/x when x>1, f(x) will ALWAYS be greater than g(x) [where both are defined] and therefore cannot be equal to each other

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  • $\begingroup$ I am sorry but this is a low quality answer on two points of view 1) Mathematicaly, it is very hard to follow you (for example your use of "rate of change" that btw should be called a "derivative"). 2) Moreover you must use Latex for your formulas ! $\endgroup$ – Jean Marie Apr 1 '17 at 6:18
  • $\begingroup$ @JeanMarie On the one hand, he's young. But on the other hand, he's been around this site long enough to know $\LaTeX$. $\endgroup$ – David R. Apr 1 '17 at 18:40
  • $\begingroup$ Hint: Type -\infty to get $-\infty$. $\endgroup$ – David R. Apr 1 '17 at 18:40
  • $\begingroup$ @David R. You are right. I had not a look at the information "eighth grader". Taking account of this level, I can withdraw my first remark. $\endgroup$ – Jean Marie Apr 1 '17 at 20:50
  • $\begingroup$ I apologize for my hasty answer. I'll try to spend more time on the formatting next time. @JeanMarie $\endgroup$ – Saketh Malyala Apr 1 '17 at 21:04

Plot of $\color{blue}{y=2x}$ and $y(x)=\log x$.


There is no solution in $\mathbb{R}$.

The complex plane

Given $z=x+i y$, the logarithm $$ \log z = \log |z| + i \arg z $$ To solve, demand equality between both the real and imaginary parts. $$ \begin{align} \log z &= 2z \\ \log |z| &= 2x \\ \arg z &= y \end{align} $$

Below are the level surfaces for $$ \log z - 2z = 0; $$ the real part is on the left, the imaginary on the right.


Is there a solution? Is there a point where the $0$ contours cross? No. Below, the $0$ contours are shown on the same plot.

enter image description here

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  • 1
    $\begingroup$ The conclusion is correct, but a plot is not a proof. $\endgroup$ – dxiv Apr 1 '17 at 4:31
  • $\begingroup$ when I put this up on wolfram alpha it gives, $\endgroup$ – satan 29 Apr 1 '17 at 4:36
  • $\begingroup$ an imaginary solution... $\endgroup$ – satan 29 Apr 1 '17 at 4:36
  • $\begingroup$ actually, the problem was evaluating 100^100^100.... $\endgroup$ – satan 29 Apr 1 '17 at 4:40
  • $\begingroup$ @satan 29: Do you want to include complex solutions? $\endgroup$ – dantopa Apr 1 '17 at 4:40

(I use below notation $\ln$ instead of $\log$).

We are going to prove that

$$\tag{1}\forall x>0, \ \ \ \ \ln(x) \ \ \leq \ \ 2x-a \ \ < \ \ 2x \ \ \ \text{with} \ \ \ a:=(1+\ln(2))>0$$

(see picture below). As a consequence of (1), equation $\ln(x)=2x$ has no (real) solution.

We have only to prove the first inequality in (1) (the second one is immediate).

In fact, it suffices to prove that $y=2x-a$ is the equation of a certain tangent, because, $\ln$ being a concave function, its curve is always situated below any of its tangents.

The general equation of the tangent in $x_0$ to the curve with cartesian equation $y=f(x)$ is


It becomes here, for $f=\ln$ and $x_0=\tfrac12$:

$y=-\ln(2)+\dfrac{1}{1/2}\left(x-\frac12\right)$, i.e., $y=2x-(1+\ln(2)),$

ending the proof.

enter image description here

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