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

1 Answer

3 votes

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}

User Marc Cayuela
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