Solve for X? (Logarithmic Equation)

#log_9(x-5) + log_9(x+3) = 1#

2 Answers
Apr 17, 2017

#x=6# and #x = -4#

Explanation:

Recall that #log_{a} AB = log_{a}A + log_{a} B# and #log_{a}a = 1.#
Therefore, # log_{9}(x-5) + log_{9}(x+3) = 1# can be written as #log_{9}{(x-5)(x+3)}=log_{9}9.#
Cancel out #log_{9}# on both sides, we have #(x-5)(x+3)=9#; and solving #(x-5)(x+3)-9=0# for #x# we have #x=6, -4#

Apr 17, 2017

#x=6#

Explanation:

Using properties of logarithms we can rewrite the left hand side.

#log(a)+log(b)=log(ab)#

#log_9((x-5)(x+3))=1#

Now rewrite both sides in terms of the base #9#

#9^(log_9((x-5)(x+3)))=9^1#

rewriting the left hand side we have

#(x-5)(x+3)=9#

#x^2-2x+15=9#

#x^2-2x-24=0#

#(x-6)(x+4)=0#

#x-6=0# OR #x+4=0#

#x=6#

If we are restricted to the real numbers, we disregard
#x=-4#