Question

If $32^{a}$ = $\frac{1}{256^b}}$ , then find 5a + 8b

Answer 1

Answer:

$32 {}^{a} = \frac{1}{256 {}^{b} } \\ 2 {}^{5 \times a} = \frac{1}{2 {}^{8 \times b} } \\ 2 {}^{5a} = \frac{1}{2 {}^{8b} } \\ 2 {}^{5a} = 2 {}^{ - 8b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{x {}^{y} } = x {}^{ - y}$

I hope it will help you.

Now here base is same

So

$2 {}^{5a} = 2 {}^{ - 8b} \\ 5a = - 8b \\ 5a + 8b = 0$