Prove that $2^x-\ln(x)=x$ iff $x$ is an odd integer?

Question

Here's what I have so far:

Since $x$ is an odd integer, $x=2b+1$, $b \in \mathbb{Z}$. I plug that in for $x$, so $2^{(2b+1)}-\ln(2b+1)=2b+1$.

I raise $e$ on both sides, so $e^{2(2b+1)-\ln(2b+1)}=e^{2b+1}$. Then I simplify to $e^{2^{(2b+1)}}-e^{2b+1}=2b+1$. Finally, I get $e^{2b+1}=2b+1$. I'm not sure how to solve for b from here. Any hints?

Answer 1

The only thing we need to do, when we wish to show that a statement is false, is to give a single counter-example. Let $x = 1$ (which is an odd integer), in the equation $$\color{red}{2^x - \log x} = \color{blue}{x} $$ then we have $$\color{red}{2^1 - \ln 1} = \color{red}{2 - 0} = \color{red}{2}\neq \color{blue}{1}.$$ We have now given a single counter-example, and hence the claim has been proven false.

Answer 2

It's extremely false. Consider the gradient of $f(x) = 2^x - \log(x) - x$; we'll show that $f$ is increasing past $x=2$, and also that $f(2) > 0$, and so $f(x) > 0$ for all $x > 2$.

We have $$f'(x) = \log(2) \times 2^x - \frac{1}{x} - 1$$

$f'$ is strictly increasing (differentiate it to see that $f'$ is strictly positive in gradient). Moreover, $f'$ is positive at $x=2$, since $4 \log(2) - \frac{1}{2} - 1 \simeq 1.27$, and it only gets more positive as $x$ increases. Hence in particular $f$ is increasing (and, indeed, increasing more and more rapidly) as $x$ increases beyond $2$.

And $f$ is also positive at $x=2$ since $2 - \log(2) \simeq 1.3$.

Therefore $f$ never hits $0$ past $x=2$.

So in particular $2^x - \log(x) \not = x$ for $x > 2$. (It so happens that it's never $0$ anywhere.)

Answer 3

This is false because $\ln x$ is trascendental when $x$ is integer and $2^x-x$ would be an integer.