Solve $\log_9 (a) + \log_{12} (b) = \log_{16} (a+b)$ for $a/b$ The question:
$$\log_9 (a) + \log_{12} (b) = \log_{16} (a+b)$$
solve for $a/b$.
It gives hints:
put it all in terms of x.
$$9^x=a$$
$$12^x=b$$
$$16^x=a+b$$
Now prove that:
$b^2=a(a+b)$ I did and it does equal.
Then it tells me to divide both sides of $b^2=a(a+b)$ by $b^2$. Here is what I got:
$$\frac{b^2}{b^2} = \frac{a(a+b)}{b^2}$$
$$1 = \frac{a}{b}*\frac{a+b}{b}$$
$$\frac{1}{a+b}*\frac{1}{b}=\frac{a}{b}$$
$$\frac{1}{b(a+b)}=\frac{a}{b}$$
$$\frac{1}{12^x16^x}=\frac{9^x}{12^x}$$
$$\frac{1}{192^x}=\frac{3^x}{4^x}$$
$$1=\frac{576^x}{4^x}$$
$$1=144^x$$
$$x=\log_{144}(1)=\frac{\log(1)}{\log(144)}=0$$
Therefore:
$$\frac{a}{b}=\frac{9^x}{12^x}=\frac{9^0}{12^0}=\frac{1}{1}=1$$
so $a/b = 1$ is what I got. But I looked this question up on the internet and apparently it has to be $a/b=$the golden ratio (there was a hint at the end of the revision sheet saying $a/b$ and sunflower). But I can't figure it out. I always end up with $a/b = 1$ whatever I do. Help please.

Hint : You have the following equality
$$1 = \frac{a}{b} \times \frac{a+b}{b}, $$
and observe that $\frac{a+b}{b} = \frac{a}{b}+\frac{b}{b} = \frac{a}{b} + 1$, now we define $Y=\frac{a}{b}$, to get that $Y$ is a solution of
$$1 = Y\times(Y+1)\qquad \Leftrightarrow\qquad Y^2+Y-1=0. $$
