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

1 Answer

Sep 2, 2015

# x = sqrt(11) #

Explanation:

# log a + log b = log (ab) #
# 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^2-1 = 10 #
# x = +- sqrt(11) #

Since # x > 1 #, # x = sqrt(11) #

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