Log equation with no real solutions 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.
 A: 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.

A: 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
A: (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 
$$y=f(x_0)+f'(x_0)(x-x_0)$$
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.

