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.

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.

$\begin{array}{|rcll|} \hline \log _b \left( x \right) &=& \dfrac{ \log _c \left( x \right) } { \log _c \left( b \right) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &P &=& \log_8 \left( 3 \right) \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { \log _{10} \left( 8 \right) } \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { \log _{10} \left( 2^3 \right) } \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { 3\cdot \log _{10} \left( 2 \right) } \\ \hline (1) & \log _{10} \left( 2 \right) &=& \dfrac{ \log _{10} \left( 3 \right) } {3P} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &Q &=& \log_3 \left( 5 \right) \\\\ &Q &=& \dfrac{ \log _{10} \left( 5 \right) } { \log _{10} \left( 3 \right) } \\\\ \hline (2) & \log _{10} \left( 3 \right) &=& \dfrac{ \log _{10} \left( 5 \right) } {Q} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \log _{10} \left( 5 \right) &=& \log _{10} \left( \frac{10}{2} \right) \\\\ \log _{10} \left( 5 \right) &=& \log _{10} \left( 10 \right)- \log _{10} \left( 2 \right) \quad &| \quad \log _{10} \left( 10 \right) = 1\\\\ \log _{10} \left( 5 \right) &=& 1- \log _{10} \left( 2 \right) \quad &| \quad \log _{10} \left( 2 \right) = \dfrac{ \log _{10} \left( 3 \right) } {3P}\\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \log _{10} \left( 3 \right) } {3P} \quad &| \quad \log _{10} \left( 3 \right) = \dfrac{ \log _{10} \left( 5 \right) } {Q}\\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \dfrac{ \log _{10} \left( 5 \right) } {Q} } {3P} \\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \log _{10} \left( 5 \right) } {3PQ} \\\\ \log _{10} \left( 5 \right) + \dfrac{ \log _{10} \left( 5 \right) } {3PQ} &=& 1 \\\\ \log _{10} \left( 5 \right)\cdot \left( 1+ \dfrac{ 1 } {3PQ} \right ) &=& 1 \\\\ \log _{10} \left( 5 \right)\cdot \left( \dfrac{ 3PQ+1 } {3PQ} \right ) &=& 1 \\\\ \mathbf{\log _{10} \left( 5 \right)} &\mathbf{=}& \mathbf{\dfrac{3PQ}{ 3PQ+1 } }\\ \hline \end{array}$