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

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Answer 1 · 2018-03-10T20:24:38

Use the law

$log(a) - log(b) = log(a/b)$ .

Hence

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

$(x + 1)/(x - 1) = 10$

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

$x + 1 = 10x - 10$

$11 = 9x$

$x = 11/9$

We now verify that it satisfies the equation. It will as long as $x > 1$ , which it is. Hence, we're done.

Hopefully this helps!

Answer 2 · 2018-03-10T20:25:18

The solution is $x=11/9$ .

Use these $log$ arithm rules to help simplify the equation:

$log(color(red)x)color(green)-log(color(blue)y)=log(color(green)(color(red)x/color(blue)y))$

$log(x)=log(y)qquadquad=>qquadquadx=y$

Now here's the actual problem:

$log(color(red)(x+1))color(green)-log(color(blue)(x-1))=1$

$log(color(green)(color(red)(x+1)/color(blue)(x-1)))=1$

$log(color(green)(color(red)(x+1)/color(blue)(x-1)))=log(10)$

$color(green)(color(red)(x+1)/color(blue)(x-1))=10$

$color(red)(x+1)=10(color(blue)(x-1))$

$color(red)(x+1)=10x-10$

$color(red)x+11=10x$

$11=9x$

$11/9=x$

$x=11/9$

That's the solution. Hope this helped!

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