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!