Logarithm Relations

Set $x = \log_9 a = \log_{12} b = \log_{16} (a+b)$ so that $9^x = a \;,\; 12^x = b \;, \mbox{and} \; 16^x = a+b$ . Furthermore, $b/a = (4/3)^x > 0$ . Noting that $9^x + 12^x = 16^x$ , divide by $9^x$ to get the quadratic equation $1 + (4/3)^x = [(4/3)^x]^2$ in $(4/3)^x$ . This equation has one positive solution: $(4/3)^x = \dfrac{1+\sqrt{5}}{2} = b/a$ .

In general, if $a$ and $b$ are positive numbers with $a < b$ and $p$ and $q$ are prime numbers with $p < q$ such that $\log_{p^2} a = \log_{pq} b = \log_{q^2} (a+b)$ Then $b/a = \dfrac{1+\sqrt{5}}{2}$ . The proof follows along the same lines of the solution above.

Let $\log_9 a = \log_{12} b = \log_{16}(a+b) = x$ . Therefore $a=9^x$ , $b = 12^x$ , and

$\begin{aligned} a+b & = 16^x & \small \blue{\text{Divide both sides by }a = 9^x} \\ 1 + \frac ba & = \frac {16^x}{9^x} = \frac {16^x \cdot 9^x}{9^{2x}} = \frac {12^{2x}}{9^{2x}} = \left(\frac ba\right)^2 \end{aligned}$

$\begin{aligned} \implies \left(\frac ba\right)^2 - \frac ba - 1 & = 0 \end{aligned}$

Solving the quadratic, we have $\dfrac ba = \varphi = \boxed{\frac {1+\sqrt 5}2}$ , where $\varphi$ is the golden ratio .