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

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Answer 1 · 2016-02-07T03:21:10

Since the logs are in the same base already, we can use the rule #log_ab - log_ac = log_a(b / c)#

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

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

Since this is a base 10, we can now convert to exponential form.

#((x + 1)/(x - 1)) = 10^1#

#x + 1 = 10(x - 1)#

#x + 1 = 10x - 10#

#1 + 10 = 10x - x#

#11 = 9x#

#11/9 = x#

Hopefully you understand now!