[tex]3^{x} = 4^{y} = 12^{z}[/tex]
[tex]3^{x} = 4^{y} = (4 \cdot 3)^{z}[/tex]
[tex]3^{x} = 4^{y} = 4^{z} \cdot 3^{z}[/tex]
[tex]\text{Let a } = 3^{x} = 4^{y} = 4^{z} \cdot 3^{z}[/tex]
[tex]log_3a = x[/tex]
[tex]log_4a = y[/tex]
[tex]log_{(4 \cdot 3)}a = z[/tex]
Using change of base:
[tex]x = \frac{lna}{ln3}[/tex]
[tex]y = \frac{lna}{ln4}[/tex]
[tex]z = \frac{lna}{ln(4 \cdot 3)}[/tex]
[tex]ln3 = \frac{lna}{x}[/tex]
[tex]ln4 = \frac{lna}{y}[/tex]
[tex]ln(4 \cdot 3) = \frac{lna}{z}[/tex]
Now, ln(4 · 3) = ln(4) + ln(3)
[tex]\frac{lna}{z} = \frac{lna}{x} + \frac{lna}{y}[/tex]
[tex]\frac{1}{z} = \frac{1}{x} + \frac{1}{y}[/tex]
[tex]\frac{1}{z} = \frac{x + y}{xy}[/tex]
[tex]\therefore z = \frac{xy}{x + y}[/tex]
