If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. Your answer should no longer include any logarithms.
I noted that $\log_5 10=\frac{1}{\log_{10} 5}.$
I also noted that $\log_{5} 10=\log_5 2+\log_5 5=\log_5 2+1.$ I don't know how to continue, how do I finish this problem using my strategy? 
 A: $$\log_83=P \iff \log_{2^3}3=P \iff \frac{1}{3}\log_23=P \iff \log_23=3P \iff \log_32=\frac{1}{3P}.$$
$$\log_35=Q.$$
Dividing last equalities at the end of both lines, we get: 
$$\log_52=\frac{1}{3PQ}$$
Also, we know $$\log_{10}5=\frac{1}{\log_510}=\frac{1}{\log_52\cdot5}=\frac{1}{\log_52+\log_55}=\frac{1}{\log_52+1}=\frac{1}{\frac{1}{3PQ}+1}=\frac{3PQ}{3PQ+1}$$
A: \begin{align*}
\log_8 3 = P\quad\implies \quad 3&=8^P=2^{3P}\\
\log_3 5 = Q\quad\implies \quad 5&=3^Q=(2^{3P})^Q=2^{3PQ}
\end{align*}
From the last equation it follows that $$\log_5 2=\frac1{\log_2 5}=\frac1{3PQ}.$$
A: $$
\log_{10}5 = \frac{\log_35}{\log_310} = \frac{\log_35}{\log_35 + \log_32}
= \frac{3\log_35}{3\log_35 + \log_38} = \frac{3Q}{3Q + 1/P} = \frac{3PQ}{3PQ + 1}.
$$
A: $$ \log_{10} 5 = \log_{8} 5 \times \log_{10}{8}$$
$$\log_{10}{8}=\frac{1}{\log_{8}{10}} = \frac{1}{\log_{8} 2+\log_8 5}$$
$$\log_8 5 = \log_8 3\times \log_3 5 = PQ, \log_{8} 2=\frac{1}{3}$$
$$ \log_{10} 5 = PQ\times \frac{1}{\frac{1}{3}+PQ}$$
A: An attempt. But I'm not sure whether the assumption I make at the beginning is correct, or whether it is arbitrary. 

We have : 
$P = \log_8 3 \iff 8^P = 3$
and 
$Q = log_3 5 \iff 3^Q = 5$. 
Suppose that  : $log_{10} 5 = kPQ$
$log_{10} 5 = kPQ$
$\rightarrow 10^{log_{10} 5} = 10^{kPQ}$
$\rightarrow 5 = 10^{kPQ}$
$\rightarrow 3^Q = 10^{kPQ}$
$\rightarrow {(8^P)}^Q = 10^{kPQ}$
$\rightarrow 8^{PQ} = 10^{kPQ}$
$\rightarrow 8^{PQ} = {(8^{\log_{8}{10}})}^{kPQ}$
$\rightarrow 8^{PQ} = {8^{(\log_{8}{10\times kPQ)}}}$
$\rightarrow PQ = \log_{8}{10\times kPQ}$
$\rightarrow 1 = \log_{8}{10\times k}$ ( Dividing by $PQ$ on both sides). 
$\rightarrow k = \frac {1} { \log_{8}{10}}$
Therefore,  $\log_{10}5 = k\times PQ = \frac {1} { \log_{8}{10}} \times PQ$
Symbolab gives $\frac {1} { \log_{8}{10}} \times PQ = \frac {\ln 5} {\ln 10}$
But $\frac {\ln 5} {\ln 10} = \log_{10} 5$ using the change of base formula , with base $e$. 
So, number $k$ seems to do the job. 
