How do you solve #log_2(x^2 - 10)= log_2 (3x)#?

1 Answer
Alan P.
Apr 8, 2016

#x=5#

Explanation:

#log_b p = log_b q#
#rArr p=q#

Therefore
#color(white)("XXX")log_2(x^2-10)=log_2(3x)#
#color(white)("XXX")rArr x^2-10=3x#

#color(white)("XXX")rarr x^2-3x-10=0#

#color(white)("XXX")rarr (x-5)(x+2)=0#

#color(white)("XXX")rarr x=5color(white)("XX")orcolor(white)("XX")x=-2#

But neither #log# is defined when #x=-2# and therefore this result is extraneous.

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