How do you solve #\log ( 2x - 2) + \log ( x - 6) = 2#?

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Answer 1 · 2017-07-31T12:18:20

Solution : #x =11#

#log (2x-2) + log (x-6) = 2 # or

#log {(2x-2)(x-6)} = 2 # or

# (2x-2)(x-6) = 10^2 # or

# 2x^2 -14x +12 -100 =0# or

# 2x^2 -14x - 88 =0 or x^2 - 7x - 44 =0 # or

# x^2 - 11x +4x - 44 =0 # or

#x(x-11) +4(x-11) =0 or (x-11)(x+4) # . Either

#x=11 or x = -4#. Check for #x = -4 #

#log( x-6) = log(-2) # (invalid) #:. x != -4#

Check for #x = 11 ; log 20+ log 5 = log100 =2# (Verified)

Solution : #x =11# [Ans]