Solve the equation log[x-8]^2=2-log[x+1]^2?

Question

#log[x-8]^2=2-log[x+1]^2#

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Answer 1 · 2018-04-22T04:00:48

Four solutions: #x = 9 or x=-2 or x = 1/2 (7 \pm \sqrt{41})#

# log(x-8)^2 = 2 - log(x+1)^2 = log 10^2 - log(x+1)^2 #

# log(x-8)^2 = log frac{10^2}{(x+1)^2} #

# (x-8)^2 = frac{10^2} { (x+1)^2}#

#(x-8)^2 (x+1)^2 - 10^2 = 0#

#( (x-8)(x+1) - 10 ) ((x-8)(x+1) + 10) = 0#

#(x^2 - 7x - 18)(x^2 -7x + 2) = 0#

#( x - 9)(x+ 2)(x^2 - 7x + 2) = 0#

#x = 9 or x=-2 or x = 1/2 (7 \pm \sqrt{41})#

Check:

#x=9#

#log(9-8)^2=log 1 = 0#

#2 - log(9+1)^2 = 2 - log(100) = 0 quad sqrt #

#x=-2#

#log(-2 -8)^2 = log 100 = 2#

#2 - log(-2 +1) = 2 quad sqrt #

The messy ones are harder to check. I'm going to leave them.