$3- \log_3(p) = \log_p(9)$, $p = ?$

Question

Could you please show me how to solve this problem with all the steps. I’m not sure how to do it.

$$3- \log_3(p) = \log_p(9)$$

Thanks.

Answer 1

Write $\log_p9=2\log_p3$ and $\log_p3=\frac{1}{\log_3p}=\frac 1x$

Then solve $3-x=\frac 2x$ which leads to $x=1,2$ and hence $$p=3,9$$

Answer 2

Hint: we can rewrite all the $log$'s to base $e$, which gives $3 - \frac{\ln(p)}{\ln(3)} = \frac{\ln(9)}{\ln(p)}$.

Answer 3

Using the change of base formula \begin{eqnarray*} 3- \frac{\ln p}{\ln 3} =\frac{2 \ln 3}{\ln p}. \end{eqnarray*} Now let $x=\ln p$ and we have \begin{eqnarray*} x^2-3(\ln 3 ) x +2(\ln 3)^2=0. \end{eqnarray*} Should be a doddle from here ?

Answer 4

it is equivalent to $$3-\frac{\ln(p)}{\ln(3)}=\frac{\ln(9)}{\ln(p)}$$ and multiplying by $$\ln(p)\ne 0$$ $$3\ln(p)-\frac{\ln(p)^2}{\ln(3)}=\ln(9)$$ and $$3\ln(3)\ln(p)-\ln(p)^2=2\ln(3)$$ can you finish?

Answer 5

HINT.-You have $\log_p(3)=\dfrac{\log_3(3)}{\log_3(p)}$. It follows $$X^2-3X+2=0$$ where $X=\log_3(p)$.

You get $\log_3(p)= 1\text{ or } 2$ so $p=3\text{ or } 9$.