How do you solve # log_3 x = log_3(x + 2) = log_3 2 + log_3 12#?

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Answer 1 · 2015-09-06T14:33:17

It is not obvious what the equation should be.

It seem likely that the equation should be

# log_3 x + log_3(x + 2) = log_3 2 + log_3 12#

#log_3(x(x+2)) = log_3 (2*12)# #" "# (property of logarithms)

#(x(x+2)) = (2*12)# #" "# (logarithms are one-to-one)

#x^2+2x = 24#

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

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

#x=-6# is an extraneous solution. #log_3(-6)# is not defined (in the real numbers)

#x=4# is the only solution.