$17^{2x} = 4^x$ Answer as exact value

Question

This is what I've done so far:

$\log17^{2x} = \log4^{x} \\ (2x) \log17 = (x) \log4 \\ (2x) \log17 - (x) \log4 = 0 \\ x(\log17^2+ \log4) =0$

$x = 0$

Am I doing this correctly?

Answer 1

It's $$\left(\frac{289}{4}\right)^x=1$$ or $x=0$.

Answer 2

For a different approach, take natural logs of both sides:

$$2x\ln(17)=x\ln (4)$$

Subtracting and factoring:

$$(2\ln(17) - \ln(4))x = 0,$$

which is only possible if $x=0$.

Answer 3

The aim with solving equations analytically is to write $0$ as a product of factors containing powers of $x$. So $2x(\log(17)-\log(2))=0$. Now it's clear there is only one solution, $x=0$ as $2\log(17/2)$ is a non zero constant.