Solve for $x$ when $4^{x-1} + 4^{x-3} = 272$

Question

$$4^{x-1} + 4^{x-3} = 272$$

I've tried to check if I can convert it to a logarithm but found there's not an answer ($x \in \mathbb Z$), and I'm not sure how to continue from there.

Thanks in advance for any help you provide.

Answer 1

Hint: We can factor $4^{x-1}+4^{x-3} = 4^2 \cdot 4^{x-3}+4^{x-3} = (4^2+1) \cdot 4^{x-3} = 17 \cdot 4^{x-3}$.

Also, $272$ is divisible by $17$.

Answer 2

Hint: $$4^{x-1} + 4^{x-3} = 16 (4)^{x-3} + 4^{x-3} =17 (4)^{x-3} = 272$$

Answer 3

Set $u=4^{x-1}$ and rewrite the equation as $u+\frac{u}{16}=272\to\frac{17u}{16}=272$ so $u=256\implies 4^{x-1}=256\implies x=5$