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This is an awkward question to me, it was not covered in class.

Suppose:

$$\begin{align*} \log_b 2 &= A, \\ \log_b 3 &= B, \\ \log_b 5 &= C. \end{align*} $$

Then, use the change-of-base formula to evaluate

$$\log_{b^2} 5$$

and

$$\log_{\sqrt b} 2 .$$

As an example answer, we're given $\log_3 b$ becomes $1/B$.

Hopefully someone understands where they're coming from.

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    $\begingroup$ $\log_{b^2}5=\dfrac{\log_b 5}{\log_b b^2}$; and $\log_{\sqrt b}2=\dfrac{\log_b 2}{\log_b \sqrt b}$... in general, $\log_c a=\dfrac{\log_b a}{\log_b c}$. $\endgroup$ – J. M. is a poor mathematician Nov 29 '11 at 2:24
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$\log_x y = \frac{\ln y}{\ln x}$

So,

$\log_{b^2} 5 = \frac{\ln 5}{\ln b^2} = \frac{\ln 5}{2 \ln b}$

and you should be able to continue from here.

Edit: I just realized that you may not know what $\ln$ is. In which case think of it as $\log_e$ where $e$ is a well known number $e=2.71828...$.

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$$\log_{b^2}5=\frac{\log_b5}{\log_bb^2}=\frac{\log_b5}{2}=\frac{C}{2}$$

$$\log_{\sqrt b}2=\frac{\log_b2}{\log_b\sqrt b}=\frac{\log_b2}{\frac12\log_bb}=2A$$

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