Solve $\log_2 (x - 1) + \log_2 (x + 1) = 3$.
A$x = 9$
B$x = 3$CORRECT
C$x = \pm 3$
D$x = \sqrt{7}$
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Why the answer is B, and why the others tempt you.
Combine the logarithms: $\log_2 (x-1)(x+1) = 3$, so $x^2 - 1 = 2^3 = 8$ and $x^2 = 9$. Although $x = \pm 3$ solves the quadratic, $x = -3$ makes $x - 1$ negative and the logarithm undefined, so only $x = 3$ is valid.
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