# Question #c8320

$\log \left({0.01}^{2} / {x}^{3}\right) = - 10.64$
I assume that $\log$ means the logarithm with base $10$.

By the property $\log \left(\frac{a}{b}\right) = \log \left(a\right) - \log \left(b\right)$ we get:
$\log \left({0.01}^{2}\right) - \log \left({x}^{3}\right) = - 10.64$
$\log \left({\left({10}^{- 2}\right)}^{2}\right) + 10.64 = \log \left({x}^{3}\right)$
$\log \left({10}^{- 4}\right) + 10.64 = \log \left({x}^{3}\right)$
By the property $\log \left({a}^{b}\right) = b \log \left(a\right)$
$- 4 \log \left(10\right) + 10.64 = 3 \log \left(x\right)$
$- 4 + 10.64 = 3 \log \left(x\right)$
$6.64 = 3 \log \left(x\right)$
$\frac{6.64}{3} = \log \left(x\right)$
By definition of logarithm with base $10$
$x = {10}^{\frac{6.64}{3}} \approx 163 , 43$