# How do you calculate log10^3 + log10(y^2)?

Sep 20, 2016

$\textcolor{g r e e n}{4 + {\log}_{10} {y}^{2}}$

#### Explanation:

Since the default base for $\log$ is $10$

$\log {10}^{3} = {\log}_{10} {10}^{3} = 3$

and remembering that ${\log}_{b} p q = {\log}_{b} p + {\log}_{b} q$
$\log 10 \left({y}^{2}\right) = {\log}_{10} 10 + {\log}_{10} \left({y}^{2}\right) = 1 + {\log}_{10} \left({y}^{2}\right)$

Therefore
$\log {10}^{3} + \log 10 \left({y}^{2}\right) = 4 + {\log}_{10} \left({y}^{2}\right)$