Equation for logarithmic and exponential equation

Question

I've tried to solve this by graphing, inversing the two functions, but I just can't really get the exact value.

$3^x=-\log_3{x}$

Answer 1

As said, only numerical methods would solve the problem.

By inspection, you can see that the solution is "close" to $\frac 14$. If you make a Taylor expansion around this point or (it is the same), one iteration of Newton method, you should get $$x\approx \frac 14+ \frac{\log (4)-\sqrt[4]{3} \log (3)}{4+\sqrt[4]{3} \log ^2(3)}\approx 0.239342$$ In fact, if we look at the successive iterates, we get $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.25000000000000000000 \\ 1 & 0.23934214476240469666 \\ 2 & 0.23948770249429066233 \\ 3 & 0.23948773148613359008 \\ 4 & 0.23948773148613473919 \end{array} \right)$$ which is the solution for twenty significant figures.

Answer 2

Maybe do you mean the following equation? $$x^3=\log_x3$$

This equation has a nice unique root: $\sqrt[3]3$.