Logarithm equations. Finding P

Question

Question: suppose $\mathrm{log}_9 X + \mathrm{log}_{27} X = P$. write the value of $\mathrm{log}_3 X + \mathrm{log}_{81} X$ in terms of $P$.

I changed $\mathrm{log}_9 X + \mathrm{log}_{27} X = P$ into $\frac{1}{2} \mathrm{log}_3 X + \frac{1}{3}\mathrm{log}_3 X = P$.

I can't expand $\mathrm{log}_3 X + \mathrm{log}_{81} X$ in terms of $P$.

Answer 1

Hints

Complete, justify and fill in details:

$$P=\log_9x+\log_{27}x=\frac{\log_3x}{\log_39}+\frac{\log_3x}{\log_327}=\left(\frac12+\frac13\right)\log_3x$$

Answer 2

You changed $P = \log_9(x) +\log_{27}(x) \Rightarrow (\frac{1}{2} +\frac{1}{3})\log_3(x)$. Thus we have $$\log_3(x) =\frac{6}{5}P$$. Then we have $$\log_3(x)+\log_{81}(x)= \log_3(x) +\frac{1}{4}\log_3(x) = \frac{5}{4}\log_3(x) = \frac{5}{4}\times \frac{6}{5}P = \frac{3P}{2}$$. Hope it helps.