Question

What is the true solution to the logarithmic equation?
log₂[log₂ (√4x)] = 1

Answer 1

Note that

•
`√(4x)` is defined as long as
`x\ge0`

•
`\log_2(x)` is defined as long as
`x>0`

•
`\log_2(\log_2(x))` is defined as long as

`\log_2(x) > 0 \implies 2^(\log_2(x)) > 2^0 \implies x > 1`

so that overall, any solution to this equation must be larger than 1.

In the equation itself, introduce powers of 2 and squares to eliminate the logarithms and square root. Solve for
`x`.

`\log_2\left(\log_2\left(√(4x)\right)\right) = 1`

`2^(\log_2\left(\log_2\left(√(4x)\right)\right)) = 2^1`

`\log_2\left(√(4x)\right) = 2`

`2^(\log_2\left(√(4x)\right)) = 2^2`

`√(4x) = 4`

`\left(√(4x)\right)^2 = 4^2`

`4x = 16`

`\boxed{x = 4}`