How do you solve #log(x-3)+log(x+5)=log9#?

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Answer 1 · 2017-01-04T01:55:36

# x=4#, and possibly #x=-6#, depending upon the context of the question and what led to the logarithmic equarion

#log(x-3)+log(x+5)=log9#

Using #logAB=logA+logB# we get:

#log(x-3)(x+5)=log9#

Using #log A=log B <=> A=B# we have

#(x-3)(x+5)=9#
# :. x^2+5x-3x-15=9#
# :. x^2+2x-24=0#
# :. (x+6)(x-4)=0#
# :. x=4,-6#

Now depending upon the context of the question we could probably eliminate #x=-6#.

#x=-6#, in the original form of the question, would lead to the log of negative numbers which do not exist.