# What is the value of x in log(x+ 2)!- log(x + 1)! = 2?

Dec 18, 2016

$x = 98$

#### Explanation:

Use the rule color(magenta)(log_a n - log_a m = log_a (n/m).

log ((x + 2)!)/((x + 1)!) = 2

Expand the factorials using the definition color(magenta)(n! = n(n - 1)(n - 2)(...)(1)

((x + 2)(x + 1)!)/((x+ 1)!) = 10^2

Eliminate the factorials, to be left with:

$x + 2 = 100$

$x = 98$

Hopefully this helps!