# log ( x + 1 ) / log ( x - 1 ) = 2# ?

1 Answer
P dilip_k
Feb 23, 2018

x=3

Explanation:

# log ( x + 1 ) / log ( x - 1 ) = 2#

#=> log_10 ( x + 1 ) / log_10 ( x - 1 ) = 2#

#=> log_10 ( x + 1 ) xx log_ "(x - 1 )" 10 = 2#

#=> log_ "(x - 1 )" (x+1) = 2#

#=>(x-1)^2=x+1#

#=>x^2-3x=0#

#=>x(x-3)=0#

#=>x=0,3#

For #x=0, log_10(x-1) "is not valid as "log_10(-1) " not feasible"#

So #x=3#

Checking validity for x=3

# LHS=log ( x + 1 ) / log ( x - 1 ) = log ( 3 + 1 ) / log ( 3 - 1 )#

#=(log4)/log2=(2log2)/log2=2=RHS#

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