Question pls

Note that $\large{0.125 = {1 \over 8}}$. This can be written as $\large{1 \over 2^3}$ or $\large{2^{-3}}$

We now have $(2^{-3})^3$. We can simplify this by multiplying the exponents. 

Can you solve it now?

$(\frac{1}{8})^3 = \frac{1}{512} = 2^a \ |\cdot 512 \\ 1 = 2^{a+9}$

We can always express 1 as n² ! Let n be 2:

$2^0 = 2^{a+9}$

because if n^{m + a} = n^{d + k}, then it applies m + a = d + k:

 0 = a + 9 | - 9

-9 = a

a = -9

That's the only possible solution.

Thank you!!