Is it possible to find the value of $x$ where $e^x$ exceeds $x^{10}$ by hand?

Question

All I managed is to "simplify" the equation $e^x=x^{10}$ to $\frac{x}{\ln{x}}=10$. Is there some way or trick to make the equation look like $x=\dots$? (Solve the equation, in other words.)

Answer 1

I presume you want the positive solution. Write the equation as $f(x) = 0$ where $f(x) = e^{x/10} - x$. A good initial guess is $x_0 = 1$. Newton's method is the iteration $$ x_{n+1} = x - \dfrac{f(x)}{f'(x)} = \dfrac{e^{x/10} (10 - x)}{10 - e^{x/10}}$$ You get $x_1 \approx 1.118238267$, $x_2 \approx 1.118325592$, and $x_3$ is the same (to 10 significant digits). So the solution is approximately $1.118325592$.

Answer 2

Even more simply:

You want to find $x$ such that $x = e^{x/10}$. For small $x$, $e^{x/10} \sim 1+x/10 $, so, approximately, $x \sim 1+x/10$ or $x \sim 10/9 = 1.111...$.

If $f(x) = e^{x/10}$, $f'(x) =e^{x/10}/10 \sim (1.111...)/10 < 1 $, so the iteration $ x \gets f(x) $ should converge.

And the iterations, starting at $10/9$, are $1.1175190687, 1.1182353997, 1.1183155052, 1.1183244636, 1.1183254654 $ and it is, as expected, converging.

Of course Newton's method, as in Robert Israel's answer, converges faster. (From what I have seen, his mind converges much faster than mine, also.)