What is the true solution to the logarithmic equation?

log2 [log2 (sqrt4x) ]=1

Answer:4

Step-by-step explanation:

log₂[log₂ (√4x) ] = 1

log₂2 = 1

So we replace our 1 with log₂2

log₂[log₂ (√4x) ] = log₂2

log₂ on bothside will cancel each other.

We will be left with;

[log₂ (√4x) ] = 2

log = power of exponential

√4x = 2²

√4x = 4

Square bothside

(√4x) ² = 4²

4X = 16

Divide bothside by 4

4x/4 = 16/4

x = 4