# Proof an equation is wrong?

How can I prove that the equation $r=16^r$ is wrong for any arbitrary value of $r$? I have tried:

\begin{align} &r=16^r &&\implies \log_r r = \log_r 16^r \\ &&& \implies 1 = r\log_r 16 \\ &&& \implies 1/r = \log_r 16 \\ &&& \implies r^{1/r} = r^{\log_r 16}\\ &&& \implies \sqrt[r]{r} = 16 \end{align} I am stuck here.

• I'm not sure what the problem is. Are you trying to show that there aren't any solutions to $r = 16^r$? – Cameron Williams Jul 20 '14 at 19:55
• You could differentiate $f(r) = 16^r - r$ to locate the minimum, and then check that the difference is always positive. – Daniel Fischer Jul 20 '14 at 20:01
• @CameronWilliams yes I am trying to show that $r=16^r$ has no solutions. – Sharod Saor Jul 20 '14 at 20:23

I'm not too good with number theory, so I went about this in a different way and took a calculus-based approach.

Consider the function $f:(0,\infty) \to \mathbb{R}$ defined by $f(x) = x^{1/x}$. Then $$f'(x) = -x^{\frac{1}{x} - 2}(ln(x) - 1).$$

Then $f'(x) > 0$ for $x \in (0, e)$ and $f'(x) < 0$ for $x \in (e, \infty)$. Thus $f$ attains a global maximum at $x = e$.

So for all $x$, $$f(x) \leq f(e) = e^{1/e} \approx 1.445 < 16.$$ Thus, there exists no $r$ s.t. $r^{1/r} = 16$.

$16^{r}>0$ for all r and so, for $r=16^{r}$ to hold, you must have $r>0$.

Then observe from the Taylor expansion: $$16^r-r=1+\underbrace{r(\log (16)-1)}_{>0}+\sum_{k=2}^\infty \underbrace{\frac{\log^k(16)}{k!}r^k}_{>0} > 0$$ for all $r>0$.

• This generalizes to $b^r > r$ if $b>e$. – Semiclassical Jul 20 '14 at 20:26

Draw a graph of y=x and y=16^x on the same axes, then I'm sure you can find a reason why there can't be a solution