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

Question

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

1 Answer

Impact of this question

1873 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-02T16:13:11

$x = \frac{11}{9}$ $x = 11/9$

Explanation:

$log a - log b = log (\frac{a}{b})$ $log a - log b = log(a/b)$
${log}_{a} b = c \Leftrightarrow a^{c} = b$ $log_a b = c <=> a^c = b$

$log (x + 1) - log (x - 1) = 1$ $log (x+1) - log (x-1) = 1$
$\frac{log (x + 1)}{x - 1} = 1$ $log (x+1)/(x-1) = 1$
Assuming $log$ $log$ base 10:
$\frac{x + 1}{x - 1} = 10$ $(x+1)/(x-1) = 10$
$x + 1 = 10 (x - 1)$ $x+1 = 10(x-1)$
$x = \frac{11}{9}$ $x = 11/9$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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