Given [tex] p,q>0 [/tex] and [tex] \log_9p=\log_{12}q=\log_{16}(p+q)=x [/tex] (say).
Then,
[tex] p=9^x\\
q=12^x\\
p+q=16^x [/tex]
From the above 3 equations,
[tex] \frac{q}{p} =(\frac{12}{9} )^x\\
\frac{q}{p} =(\frac{4}{3} )^x\\
\frac{p+q}{p} =(\frac{16}{9} )^x\\
\frac{p+q}{p} =(\frac{4}{3} )^{2x}\\ [/tex]
From the equations, we get
[tex] \frac{p+q}{p}=(\frac{q}{p})^2\\
1+\frac{q}{p}=(\frac{q}{p})^2\\
(\frac{q}{p})^2-\frac{q}{p}-1=0\\
\frac{q}{p}=\frac{1 \pm \sqrt{5}}{2} [/tex]
Since [tex] p,q>0 [/tex], the negative value is rejected.
[tex] \frac{q}{p}=\frac{1 + \sqrt{5}}{2} [/tex]
