# Quotient of Arguments to a Logarithm The problem I'm working on is: There are p , q &gt;

Quotient of Arguments to a Logarithm
The problem I'm working on is:
There are $p,q>0$ satisfying ${\mathrm{log}}_{9}p={\mathrm{log}}_{12}q={\mathrm{log}}_{16}\left(p+q\right)$. Find $q/p$
I set all of the logarithms equal to a variable t, so I could say
${9}^{t}=p,{12}^{t}=q,{16}^{t}=p+q$
which implies
${9}^{t}+{12}^{t}={16}^{t}.$
I also know that
$\frac{q}{p}=\frac{{12}^{t}}{{9}^{t}}={\left(\frac{4}{3}\right)}^{t}$
$\frac{p+q}{q}=\frac{{16}^{t}}{{12}^{t}}={\left(\frac{4}{3}\right)}^{t}$
and therefore
$\frac{q}{p}=\frac{p+q}{q}.$
I haven't gotten any farther than this, so does anyone have advice?
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Jayvion Tyler
We have $9\cdot 16=144={12}^{2}$ . Thus ${q}^{2}=p\left(p+q\right)$. Rewrite as
${q}^{2}-pq-{p}^{2}=0.$
Divide through by ${p}^{2}$. We get
${\left(\frac{q}{p}\right)}^{2}-\frac{q}{p}-1=0.$
Solve this quadratic equation for $\frac{q}{p}$, using the Quadratic Formula.
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Gretchen Schwartz
Notice that $\frac{p+q}{q}=\frac{p}{q}+1=\frac{1}{q/p}+1$. Substituting $x=\frac{q}{p}$ into your equation, you thus get
$x=\frac{1}{x}+1$
and multiplying through by x gives you ${x}^{2}=1+x$ . This is a quadratic equation, which you can solve for x (which is equal to $\frac{q}{p}$.)
Keep in mind that it is specified that $p,q>0$