If $log_{9}x=log_{12}y=log_{16}(x+y)$ then find $y/x$.
I simplified this into
$\frac { \log { x } }{ \log { 9 } } =\frac { \log { y } }{ \log { 12 } } =\frac { log(x+y) }{ \log { 16 } } $
I equated two terms to get three equations. But that didn't give the answer. So when I looked at its step by step solution then I found
$\frac { \log { x } }{ \log { 9 } } =\frac { \log { y } }{ \log { 12 } } =\frac { log(x+y) }{ \log { 16 } } =\frac { log\frac { y }{ x } }{ log\frac { 12 }{ 9 } } =a$
After this step I understood the solution but I don't how the last term equal(i.e $\frac { log\frac { y }{ x } }{ log\frac { 12 }{ 9 } } $) equal to others terms.
Any help would be appreciated. Please help.
If $\frac{\log x}{\log 9} = \frac{\log y}{\log 12}$, then $\frac{\log x}{\log 9} - \frac{\log y}{\log 12} = 0$, i.e. $\frac{\log 12 \cdot \log x - \log 9\cdot \log y}{\log 9\cdot \log 12} = 0$ and thus $$ \log 12 \cdot \log x - \log 9\cdot \log y = 0\,. $$ Rewriting, $$\begin{align*} 0 &= \frac{\log 12\cdot \log x}{\log 12 - \log 9} - \frac{\log 9\cdot \log y}{\log 12 - \log 9} = \log x + \frac{\log 9\cdot \log x}{\log 12 - \log 9} - \frac{\log 9\cdot \log y}{\log 12 - \log 9} \\ &= \log x + \log 9\cdot \frac{\log x - \log y}{\log 12 - \log 9} = \log x + \log 9\cdot \frac{\log\frac{x}{y}}{\log\frac{12}{9}}. \end{align*}$$
Rearranging the terms, $$ -\frac{\log\frac{x}{y}}{\log\frac{12}{9}} = \frac{\log x}{\log 9} $$ that is $$ \frac{\log\frac{y}{x}}{\log\frac{12}{9}} = \frac{\log x}{\log 9} . $$