# log 4 (x) = sqrt(log 4 (x)) find x values

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### 3 Answers

log 4 (x) = sqrt(log 4 (x))

Let us square both sides :

==> (log 4 (x))^2 = log 4 (x)

Now move log 4 (x) to the left side:

==>( log 4 (x))^2 - log 4 (x) = 0

Now factor log 4 (x):

==> log 4 (x) [ (log 4 (x)) -1 ] =0

==> log 4 (x) = 0 ==> x1 = 1

and log 4 (x) -1 = 0 ==> log 4 (x) = 1 ==> x2 = 4

**Then x = {1,4}**

The first step is to eliminate the square root, and for this reason, we'll raise to square, both sides of the equation:

[log 4 (x)]^2 = [sqrt(log 4 (x))]^2

[log 4 (x)]^2 = log 4 (x)

We'll subtract log 4 (x) both sides:

[log 4 (x)]^2 - log 4 (x) = 0

We'll use the substitution technique to solve the equation:

log 4 (x) = t

t^2 - t = 0

We'll factorize:

t(t-1) = 0

t1 = 0

t-1 = 0

t2 = 1

log 4 (x) = 0

x = 4^0

x = 1

log 4 (x) = 1

x = 4^1

x = 4

**The solutions of the equation are: x1 = 1 and x2 = 4.**

log4(x) = sqrt(log4(x)) . To find x.

Solution:

log 4 (x) = sqrt (log4(x)) squaring both sides, we get:

(log 4 (x) )^2 = log4(x). . Or

{log4x)^2 -log4(x) }= 0

log4(x) { log4(x) - 1} = 0.

Therfore log4 (x) = 0. Or log4 (x) -1 = 0.

log 4 (x) = 0 gives by definition , x =4^0 = 1

log4 (x) = 1 gives by definition, x = 4^1 =4.