# How do you solve 2logx=1.6?

May 26, 2016

i found: $x = {b}^{0.8}$ where $b$ is the base of your log.

#### Explanation:

It depends on the log's base but we can say that is, say, $b$, so we have:
$2 {\log}_{b} \left(x\right) = 1.6$
rearranging:
$2 {\log}_{b} \left(x\right) = \frac{16}{10}$
${\log}_{b} \left(x\right) = \frac{1}{2} \cdot \frac{16}{10}$
${\log}_{b} \left(x\right) = \frac{1}{\cancel{2}} \cdot {\cancel{16}}^{8} / 10 = 0.8$

using the definition of log:
$x = {b}^{0.8}$

Now you can use the appropriate $b$ for our case:

if $b = 10$, for example:

$x = {10}^{0.8} = 6.30957$