# Question #f472a

##### 1 Answer
Jun 9, 2015

$z = \frac{y}{3 \frac{y}{x} + \frac{3}{2}}$

#### Explanation:

First, you want to turn the $x$, $y$ and $z$ powers into factors. To do so, use the naperian logarithm:

$\textcolor{g r a y}{N o t e : \ln \left(a \cdot b\right) = \ln \left(a\right) + \ln \left(b\right) \mathmr{and} \ln \left({a}^{b}\right) = b \ln \left(a\right)}$

$\ln \left({2}^{x}\right) = \ln \left({3}^{y}\right) = \ln \left({\left(24 \sqrt{3}\right)}^{z}\right)$

$= x \ln \left(2\right) = y \ln \left(3\right) = z \ln \left(24 \sqrt{3}\right)$

Then, isolate $z$:

$z = \frac{y \ln \left(3\right)}{\ln \left(24 \sqrt{3}\right)} = \frac{y \ln \left(3\right)}{\ln \left(3 \cdot {2}^{3}\right) + \ln \left({3}^{\frac{1}{2}}\right)}$

$= \frac{y \ln \left(3\right)}{3 \ln \left(2\right) + \ln \left(3\right) + \ln \frac{3}{2}}$

Replace $\ln \left(2\right)$ with $\frac{y}{x} \ln \left(3\right)$

$z = \frac{y \ln \left(3\right)}{3 \frac{y}{x} \ln \left(3\right) + \ln \left(3\right) + \ln \frac{3}{2}}$

$= \frac{y \cancel{\ln \left(3\right)}}{\left(3 \frac{y}{x} + 1 + \frac{1}{2}\right) \cancel{\ln \left(3\right)}} = \frac{y}{3 \frac{y}{x} + \frac{3}{2}}$