log 2 x - log 4 x = 2 find x

The equation `log_2 x - log_ 4 x = 2` has to be solved for
x.

First convert all the logarithm to a common base. As the
numbers here are 2 and powers to two we use the base 2. Use the following property
useful when changing the base of logarithms.

`(log_b x) =
(log_n x)/(log_n b)`

`log_2 x - log_ 4 x =
2`

`log_2 x - (log_2 x)/(log_2 4) =
2`

`log_2 x - (log_2 x)/(log_2 2^2) =
2`

use the property `log a^b = b*log
a`

`log_2 x - (log_2 x)/(2*log_2 2) =
2`

Now `log_b b = 1`

`log_2 x
- (log_2 x)/(2*1) = 2`

`(1/2)*log_2 x = 2`

`log_2 x = 4`

If `log_b x =
y` , `x = b^y`

`x = 4^2`

x =
16

The required solution is x = 16