How do you evaluate the expression #log_11(11^(n-5))#?

Question

1511 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-17T14:55:00

The expression can be simplified to #n - 5#.

Start by simplifying the expression using the rule #logn^a = alogn#.

#=(n - 5)log_11(11)#

Now, simplify using the formula #log_a(n) = logn/loga#.

#=(n - 5)log11/log11#

#=(n - 5)(1)#

#= n - 5#

Hopefully this helps!