How do you solve #2 log_4 x =1+ log_4 (x + 8)#?

1 Answer
Noah G
May 7, 2016

Solve using the following properties:
• #alogn = logn^a#
• #log_a(n) - log_a(m) = log_a(n/m)#

Explanation:

#log_4(x^2) - log_4(x + 8) = 1#

#log_4((x^2)/(x + 8)) = 1#

#(x^2)/(x + 8) = 4#

#x^2 = 4(x + 8)#

#x^2 = 4x + 32#

#x^2 - 4x - 32 = 0#

#(x - 8)(x + 4) = 0#

#x = 8 and x = -4#

Checking in the original equation, we find that only #x = 8# works, so the #x = -4# is extraneous and you therefore will not include it inside the solution set.

Hopefully this helps!

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