How do you solve $log(16+2b)=log(b^2-4b)$ ?

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Answer 1 · 2016-08-22T22:21:43

${-2, 8}$ is the solution set.

Use the property $log(a) = log(b) -> a = b$

$16 + 2b = b^2 - 4b$

$0 = b^2 - 6b - 16$

$0 = (b - 8)(b +2)$

$b = 8 and -2$

Checking in the original equation, both solutions work, hence the solution set of ${-2, 8}$ .

Hopefully this helps!

How do you solve log(16+2b)=log(b^2-4b)log(16+2b)=log(b^2-4b)?