How do you solve #log_3 (x + 3) + log_3 (x + 5) = 1#?

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Answer 1 · 2015-12-06T06:51:27

x=-2

#log(base3)(x+3)+log(base 3)(x+5)=1#-> use product rule of logarithm

log(base3)((x+3)(x+5))=1 write in exponential form

#3^1=(x+3)(x+5)#
#x^2+8x+15=3#
#x^2+8x+12=0#
#(x+6)(x+2)=0#

#x+6=0 or x+2=0#

x=-6 or x=-2

x=-6 is extraneous . An extraneous solution is root of transformed but it is not a root of original equation.

so x=-2 is the solution.