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

Question

15974 views around the world

You can reuse this answer
Creative Commons License

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!

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