How do you solve $log_2(3x + 1) + log_2(x + 7) = 5$ and find any extraneous solutions?

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Answer 1 · 2016-09-27T13:39:24

$log_2((3x+ 1)(x + 7)) = 5$

$(3x + 1)(x + 7) = 2^5$

$3x^2 + x + 21x + 7 = 32$

$3x^2 + 22x - 25 = 0$

$3x^2 - 3x + 25x - 25 =0$

$3x(x - 1) + 25(x - 1) = 0$

$(3x + 25)(x- 1) = 0$

$x = -25/3 and 1$

However, $x = -25/3$ renders the original equation undefined, so the only solution is $x = 1$ .

Hopefully this helps!

How do you solve log_2(3x + 1) + log_2(x + 7) = 5 log_2(3x + 1) + log_2(x + 7) = 5 and find any extraneous solutions?