Question

Given that 3^x = 4^y = 12^z, show that z = (xy)/(x+y).

Answer 1

$3^{x} = 4^{y} = 12^{z}$
$3^{x} = 4^{y} = (4 \cdot 3)^{z}$
$3^{x} = 4^{y} = 4^{z} \cdot 3^{z}$

$\text{Let a } = 3^{x} = 4^{y} = 4^{z} \cdot 3^{z}$
$log_3a = x$
$log_4a = y$
$log_{(4 \cdot 3)}a = z$

Using change of base:
$x = \frac{lna}{ln3}$
$y = \frac{lna}{ln4}$
$z = \frac{lna}{ln(4 \cdot 3)}$

$ln3 = \frac{lna}{x}$
$ln4 = \frac{lna}{y}$
$ln(4 \cdot 3) = \frac{lna}{z}$

Now, ln(4 · 3) = ln(4) + ln(3)

$\frac{lna}{z} = \frac{lna}{x} + \frac{lna}{y}$
$\frac{1}{z} = \frac{1}{x} + \frac{1}{y}$
$\frac{1}{z} = \frac{x + y}{xy}$

$\therefore z = \frac{xy}{x + y}$