How do you solve #log (x^2-9) = log (5x+5)#?

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Answer 1 · 2015-08-01T18:54:02

#color(red)(x=7)#

#log(x^2-9) = log (5x+5)#

Convert the logarithmic equation to an exponential equation.

#10^( log(x^2-9)) = 10^( log (5x+5))#

Remember that #10^logx =x#, so

#x^2-9 = 5x+5#

Move all terms to the left hand side.

#x^2-9-5x-5 = 0#

Combine like terms.

#x^2-5x-14 = 0#

Factor.

#(x-7)(x+2) = 0#

#x-7 = 0# and #x+2=0#

#x=7# and #x=-2#

Check:

#log(x^2-9) = log (5x+5)#

If #x=7#

#log(7^2-9) = log (5(7)+5)#

#log(49-9) = log (35+5)#

#log40 = log40#

#x=7# is a solution.

If #x=-2#,

#log((-2)^2-9) = log (5(-2)+5)#

#log(4-9) = log (-10+5)#

#log(-5) = log (-5)#

#log(-5) is not defined,

#x=-2# is a spurious solution.