How do you solve #\log _ { 10} ( x - \pi ) = 5#?
1 Answer
Mar 31, 2017
Explanation:
Okay, so we know that the base of a log is also the base of an exponent.
#a=x^b# is#log_xa=b#
Knowing this, let's write
#10^5=x-pi#
Simplify
#100000=x-pi#
Solve for
#100000+pi=x#
#rArrx=100000+pi#