1. ## Logarithmic Proof

Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

Thanks for you help!

2. Originally Posted by Stroodle
Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

Thanks for you help!
What a horrid question...

First note that $\displaystyle 12^z = (3 \cdot 4)^z = 3^z \cdot 4^z$

Focus firstly on

$\displaystyle 3^x = 12^z$

$\displaystyle 3^x = 3^z \cdot 4^z$

$\displaystyle 3^{x - z} = 4^z$

$\displaystyle \log{3^{x - z}} = \log{4^z}$

$\displaystyle (x - z)\log{3} = z\log{4}$

$\displaystyle \frac{x - z}{z} = \frac{\log{4}}{\log{3}}$.

Now focus on

$\displaystyle 4^y = 3^z \cdot 4^z$

$\displaystyle 4^{y - z} = 3^z$

$\displaystyle \log{4^{y - z}} = \log{3^z}$

$\displaystyle (y - z)\log{4} = z\log{3}$

$\displaystyle \frac{\log{4}}{\log{3}} = \frac{z}{y - z}$.

Therefore

$\displaystyle \frac{x - z}{z} = \frac{z}{y - z}$

$\displaystyle \frac{x}{z} - 1 = \frac{y}{y - z} - 1$

$\displaystyle \frac{x}{z} = \frac{y}{y - z}$

$\displaystyle \frac{y - z}{z} = \frac{y}{x}$

$\displaystyle \frac{y}{z} - 1 = \frac{y}{x}$

$\displaystyle \frac{y}{z} = \frac{y}{x} + 1$

$\displaystyle \frac{y}{z} = \frac{x + y}{x}$

$\displaystyle \frac{z}{y} = \frac{x}{x + y}$

$\displaystyle z = \frac{xy}{x + y}$.

3. Awesome. I get it now.

Thanks for your help Prove It.

4. Hello Stroodle
Originally Posted by Stroodle
Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

Thanks for you help!
Here's an alternative approach:

$\displaystyle 3^x = 12^z$

$\displaystyle \Rightarrow x\log3 = z\log12$

$\displaystyle \Rightarrow x = \frac{z\log12}{\log 3}$

Similarly $\displaystyle y = \frac{z\log12}{\log4}$

$\displaystyle \Rightarrow \frac{xy}{x+y}= \frac{\dfrac{z\log12}{\log 3}\cdot\dfrac{z\log12}{\log 4}}{\dfrac{z\log12}{\log 3}+\dfrac{z\log12}{\log 4}}$

$\displaystyle = \frac{\dfrac{z\log12}{\log 3}\cdot\dfrac{z\log12}{\log 4}}{\dfrac{z\log12}{\log 3}+\dfrac{z\log12}{\log 4}}\color{red}\times\frac{\log3\log4}{\log3\log4}$

$\displaystyle = \frac{z^2(\log12)^2}{z\log12(\log4+\log3)}$

$\displaystyle =\frac{z\log12}{\log12}$

$\displaystyle =z$

5. Originally Posted by Stroodle
Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

Thanks for you help!
HI

Or another approach .

$\displaystyle 3^x=4^y=12^z=k$

$\displaystyle 3^x=k\Rightarrow 3=k^{\frac{1}{x}}$

$\displaystyle 4^y=k\Rightarrow 4=k^{\frac{1}{y}}$

$\displaystyle 12^z=k\Rightarrow 12=k^{\frac{1}{z}}$

we know for the fact that 4 x 3 = 12

so $\displaystyle k^{\frac{1}{x}}\times k^{\frac{1}{y}}=k^{\frac{1}{z}}$

$\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$

yz+xz=xy

z(x+y)=xy

z=xy/(x+y)