How do you solve # log_10(x-1) - log_10(x+1) = 1#?

1 Answer
Feb 11, 2016

Answer:

#x=-11/9#

Please see Comments by @Lotusbluete and @Tom below for conditions under which solution exists.
Mathematically stated: The solution exists only for the equation
#log_10|x-1| -log_10|x+1|=1#

Explanation:

On the RHS use the following Basic Log rule
#log_b(m/n) = log_b(m) – log_b(n)#
and on the LHS use the equality #log_10 10=1#
#log_10((x-1)/(x+1))=log_10 10#
Taking antilog of both sides

#(x-1)/(x+1)=10#
#=>(x-1)=10(x+1)#
#=>x-1=10x+10#
#=>9x=-11#
#=>x=-11/9#