# `log_4 x*log_2 x + log_2(log_4 x)=2` 4,2,2 and 4 are in the base

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### 1 Answer

You should use the following logarithmic identity such that:

`log_a b = (log_c b)/(log_c a)`

`log_4 x = (log_2 x)/(log_2 4) => log_4 x = (log_2 x)/2`

Substituting `(log_2 x)/2` for `log_4 x ` yields:

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

You need to use the logarithmic identity `log (a/b) = log a - log b ` such that:

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

`((log_2 x)^2)/2 + log_2(log_2 x) - 1 = 2`

`((log_2 x)^2) + 2log_2(log_2 x) = 6`

`((log_2 x)^2) + log_2(log_2 x)^2 = 6`

You should come up with the following substitution such that:

`((log_2 x)^2) ` = t

`t + log_2 t = 6 => log_2 t = 6-t`

Notice that the equation `log_2 t = 6-t` is a transcendental equation, hence, you may find its solutions using the graphs of the functions `f(t) = 6-t` and `g(t) = log_2 t` such that:

Notice that the x coordinate of point of intersection of two graphs is 4, hence, you should substitute 4 for t in transcendental equation such that:

`log_24 = 6-4 => log_2(2^2) = 2 => 2log_2 2 = 2 => 2*1=2`

You need to solve for x `((log_2 x)^2) = t` such that:

`((log_2 x)^2) = 4 => (log_2 x) = +-2 => x_1 = 2^2 => x_1 = 4`

`x_2 = 2^(-2) => x_2 = 1/4`

**Hence, evaluating the solutions to the given equation yields `x_1 = 4` and `x_2 = 1/4` .**

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