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Find $\frac{b}a$ if $a$ and $b$ are positive real numbers such that $\log_9(a)=\log_{15}(b)=\log_{25}(a+2b)$.

How do I approach this? Do I necessarily need to solve for $a$ and $b$? I don't think so since the question simply asks for $\frac{b}{a}$.

Did
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user406996
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    One knows that $a=9^c$ and $b=15^c$ with $25^c=a+b$ hence $x=5^c$ and $y=3^c$ solve $x^2=y^2+xy$ and $b/a=z=x/y$ solves $z^2=1+z$, that is, since $z>0$, $z=\frac12(1+\sqrt5)$. – Did Apr 09 '17 at 20:06
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    It's $a+2b$ .. ... – Nosrati Apr 09 '17 at 20:10
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    @MyGlasses Gasp. Fortunately the same approach works, only now, $x^2=y^2+2xy$, hence $z^2=1+2z$, that is, since $z>0$, $z=1+\sqrt2$. Thanks. – Did Apr 09 '17 at 20:30

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