If $p=\log_63$ and $q=\log_65$, what is $\log_{45}12?$

Question

My try: We can write $\log_{45}12$ as $$\log_{45}12=\dfrac{\log_612}{\log_645}=\dfrac{\log_63+\log_64}{\log_65+\log_69}=\dfrac{\log_63+2\log_62}{\log_65+2\log_63}$$ How do I find $\log_62$ with $p$ and $q$?

Hint: $\log_6 6 =1$ and $6=2\times 3$. – lulu Mar 19 '23 at 13:34 — lulu, Mar 19 '23 at 13:34
$\log_6 2= \log_6 6 - \log_6 3 = 1-p\tag*{}$ – Nothing special Mar 19 '23 at 14:08 — Nothing special, Mar 19 '23 at 14:08

score 0 · Answer 1 · answered Mar 27 '23 at 08:36

Another approach: with $ \ 6^p \ = \ 3 \ $ and $ \ 6^q \ = \ 5 \ \ , \ $ we wish to find an expression for $ \ x \ $ in terms of $ \ p \ $ and $ \ q \ $ such that $ \ 45^x \ = \ 12 \ \ . \ $ We may write $$ (3^2 · 5)^x \ \ = \ \ 2^2 · 3 \ \ \Rightarrow \ \ 2^{2x} · 3^{2x} · 5^x \ \ = \ \ 2^{2x} · 2^2 · 3 \ \ \Rightarrow \ \ 6^{2x} · 5^x \ \ = \ \ \left(\frac63 \right)^{2x+2}· 3 $$ $$ \Rightarrow \ \ 6^{2x} · (6^q)^x \ \ = \ \ 6^{2x + 2} · (6^p)^{-2x - 2} · (6^p) $$ $$ \Rightarrow \ \ 6^{ qx} \ \ = \ \ 6^{ \ 2 \ + \ p \ - \ 2px \ - \ 2p} \ \ . \ $$

The resulting value for $ \ x \ $ does give us $ \ 45^x \ = \ 12 \ \ . \ $

$$ x \ = \ \frac{2 \ - \ p}{q \ + \ 2p} \ \ \approx \ \ \frac{2 \ - \ 0.6131}{0.8982 \ + \ 2·0.6131} \ \ \approx \ \ 0.6528 \ \ . $$

If $p=\log_63$ and $q=\log_65$, what is $\log_{45}12?$

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