Question

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

Answer 1

Note that

• $\sqrt{4x}$ 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 $x$.

$\log_2\left(\log_2\left(\sqrt{4x}\right)\right) = 1$

$2^{\log_2\left(\log_2\left(\sqrt{4x}\right)\right)} = 2^1$

$\log_2\left(\sqrt{4x}\right) = 2$

$2^{\log_2\left(\sqrt{4x}\right)} = 2^2$

$\sqrt{4x} = 4$

$\left(\sqrt{4x}\right)^2 = 4^2$

$4x = 16$

$\boxed{x = 4}$