# Question #8a9cf

Jul 20, 2016

$\log {2}^{x} = \frac{p}{3}$

#### Explanation:

If I understand the question properly, we have:
$\log {8}^{x} = p$

And we wish to express $\log {2}^{x}$ in terms of $p$.

The first thing we should note is that $\log {8}^{x} = x \log 8$. This follows from the following property of logs:
$\log {a}^{b} = b \log a$

Essentially, we can "bring down" the exponent and multiply it by the logarithm. Similarly, using this property on $\log {2}^{x}$, we get:
$\log {2}^{x} = x \log 2$

Our problem is now boiled down to expressing $x \log 2$ (the simplified form of $\log {2}^{x}$) in terms of $p$ (which is $x \log 8$). The central thing to realize here is that $8 = {2}^{3}$; which means $x \log 8 = x \log {2}^{3}$. And again using the property described above, $x \log {2}^{3} = 3 x \log 2$.

We have:
$p = x \log {2}^{3} = 3 x \log 2$

Expressing $x \log 2$ in terms of $p$ is now drastically easier. If we take the equation $p = 3 x \log 2$ and divide it by $3$, we get:
$\frac{p}{3} = x \log 2$

And voila - we have expressed $x \log 2$ in terms of $p$.