# Solving equations having both log and exponential forms

How can one Solve equations having both log and exponential forms: For eg...

$e^x$ $=$ $\log_{0.001}(x)$ gives $x=0.000993$

(according to wolfram-alpha http://www.wolframalpha.com/input/?i=e%5Ex%3D+log0.001%28x%29)

Thanks

• You could try using Lagrange Inversion Theorem – Simply Beautiful Art Jan 27 '16 at 23:07

First, let's change the base of the logarithm to $e$.

$$e^x=\frac{\ln(x)}{\ln(0.001)}$$

Try solving for the $x$ inside the logarithm.

$$x=e^{\ln(0.001)e^x}=e^{e^{x+\ln(\ln(0.001))}}=f(f(x))$$

Try to find the mystery function $f(x)$? If it such existed...

Then we'd have

$$x=f(f(x))$$Which has some solutions from solving the easier problem

$$x=f(x)$$

However, since I don't think there is a function $f(x)$, no closed form solution for $x$ will exist.

I do not see any way to solve this exactly. Even Wolfram-Alpha gives only a numerical solution (the other algebraic answers are just ways of transforming the question into other forms). Multiple numeric methods would work here--just choose your favorite.

• It is taking the form of: e^(-6.9e^(x))=x – NeilRoy Aug 12 '14 at 16:18

There really isn't much you can do. But here's something that might help. Let $y=e^x$. Clearly $$y=\log_{0.001}(x)$$ Which is the same as saying that $$x=0.001^y$$ And so your one equation is the same as solving the system of equations

$$\begin{cases}y=e^x\\x=0.001^y\end{cases}$$

• Yeah but i don't think it will help much since when we put back y as e^x we get x=0.001^e^x – NeilRoy Aug 12 '14 at 16:41