Dec 26, 2016

see below

## concepts applied

• $\setminus {\log}_{a} \left({b}^{c}\right) \setminus \Rightarrow c \setminus {\log}_{a} \left(b\right)$
• $\setminus {\log}_{a} \left(b\right) = x \setminus \leftrightarrow {a}^{x} = b$

if referring to the log as base 10 log, ${\log}_{10} x$:
$\frac{\setminus {\log}_{10} \left(225\right)}{\setminus {\log}_{10} \left(25\right)}$

then
numerator: $\setminus {\log}_{10} \left({15}^{2}\right) \setminus \Rightarrow 2 \setminus {\log}_{10} \left(15\right)$
denominator: $\setminus {\log}_{10} \left({5}^{2}\right) \setminus \Rightarrow 2 \setminus {\log}_{10} \left(5\right)$

becomes
$\frac{2 \setminus {\log}_{10} \left(25\right)}{2 \setminus {\log}_{10} \left(5\right)}$
the 2's cancel, which leaves you with $\frac{\setminus {\log}_{10} \left(25\right)}{\setminus {\log}_{10} \left(5\right)}$

if referring to the log as natural log, ${\log}_{e} x$ or $\ln x$:
$\frac{\setminus {\log}_{e} \left(225\right)}{\setminus {\log}_{e} \left(25\right)} \setminus \Rightarrow \frac{\setminus \ln \left(225\right)}{\setminus \ln \left(25\right)}$

then
numerator: $\setminus \ln \left({15}^{2}\right) \setminus \Rightarrow 2 \setminus \ln \left(15\right)$
denominator: $\setminus \ln \left({5}^{2}\right) \setminus \Rightarrow 2 \setminus \ln \left(5\right)$

becomes
$\frac{2 \setminus \ln \left(25\right)}{2 \setminus \ln \left(5\right)}$
the 2's cancel, which leaves you with $\frac{\setminus \ln \left(25\right)}{\setminus \ln \left(5\right)}$