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

Question

1429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T18:33:25

#x = 98#

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!