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

1 Answer

Sep 2, 2015

# x = 11/9 #

Explanation:

# log a - log b = log(a/b) #
# log_a b = c <=> a^c = b #

# log (x+1) - log (x-1) = 1 #
# log (x+1)/(x-1) = 1 #
Assuming #log# base 10:
# (x+1)/(x-1) = 10 #
# x+1 = 10(x-1) #
# x = 11/9 #

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