<s>[INST] 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? 
 [/INST] 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. 
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