How do you solve #4^x = 12#?

1 Answer
Mar 11, 2018

#x=1.79#

Explanation:

Apply the #log_4# on both sides of the equation

#log_4(4^x)=log_4(12)#

Apply an Exponent Logarithm Law:
#x* log_4(4)=log_4(12)#

#log_a(a)# always equals 1, so you get:
#x=log_4(12)#

If your calculator doesn't have log bases besides the default (10) you can apply the change of base formula:
#x=log(12)/log(4)#