# How do you solve log2sqrt(5x+1) = 4?

Sep 11, 2015

I found:
$x = 5 \times {10}^{6}$ if log in base 10
or
$x = 51$ if log in base 2

#### Explanation:

Assuming a log in base $10$ we can write:
${\log}_{10} 2 \sqrt{5 x + 1} = 4$

using the definition of log:
$2 \sqrt{5 x + 1} = {10}^{4}$
$\cancel{2} \sqrt{5 x + 1} = \frac{10000}{2}$
$\sqrt{5 x + 1} = 5000$

squaring both sides:
$5 x + 1 = {\left(5000\right)}^{2}$
$x = \frac{{\left(5000\right)}^{2} - 1}{5} \approx 5 \times {10}^{6}$

If on the other hand the base is $2$ we can write:
${\log}_{2} \sqrt{5 x + 1} = 4$
again using the definition of log we get:
$\sqrt{5 x + 1} = {2}^{4}$
$\sqrt{5 x + 1} = 16$
squaring both sides:
$5 x + 1 = 256$
$5 x = 255$
$x = 51$