Answer
(a) $\frac{\log_2x}{\log_3x}=\log_23$
(b) $\frac{\log_2x}{\log_8x}=3$
(c) $\frac{\log_xa }{\log_{x^2}a}=2$
Work Step by Step
*Recall the Change of Base Formula: $$\log_ax=\frac{\ln x}{\ln a}$$
(a) $$\frac{\log_2x}{\log_3x}$$
Apply the Change of Base Formula here:
$\log_2x=\frac{\ln x}{\ln 2}$ and $\log_3x=\frac{\ln x}{\ln 3}$
Therefore, $$\frac{\log_2x}{\log_3x}=\frac{\frac{\ln x}{\ln 2}}{\frac{\ln x}{\ln3}}=\frac{\ln x}{\ln 2}\times\frac{\ln3}{\ln x}=\frac{\ln 3}{\ln2}$$
Here we can apply the Change of Base Formula one more time, meaning $\frac{\ln 3}{\ln2}=\log_23$
In conclusion: $$\frac{\log_2x}{\log_3x}=\log_23$$
(b) $$\frac{\log_2x}{\log_8x}$$
Apply the Change of Base Formula here:
$\log_2x=\frac{\ln x}{\ln 2}$ and $\log_8x=\frac{\ln x}{\ln 8}$
Therefore, $$\frac{\log_2x}{\log_8x}=\frac{\frac{\ln x}{\ln 2}}{\frac{\ln x}{\ln8}}=\frac{\ln x}{\ln 2}\times\frac{\ln8}{\ln x}=\frac{\ln 8}{\ln2}=\frac{\ln2^3}{\ln2}$$
Apply Power Rule: $$\frac{\log_2x}{\log_8x}=\frac{3\ln2}{\ln2}=3$$
(c) $$\frac{\log_xa }{\log_{x^2}a}$$
Apply the Change of Base Formula here:
$\log_xa=\frac{\ln a}{\ln x}$ and $\log_{x^2}a=\frac{\ln a}{\ln x^2}$
Therefore, $$\frac{\log_xa}{\log_{x^2}a}=\frac{\frac{\ln a}{\ln x}}{\frac{\ln a}{\ln x^2}}=\frac{\ln a}{\ln x}\times\frac{\ln x^2}{\ln a}=\frac{\ln x^2}{\ln x}$$
Apply Power Rule: $$\frac{\log_xa}{\log_{x^2}a}=\frac{2\ln x}{\ln x}=2$$
