How do you solve #4^(2x+3)=1#?

1 Answer
Aug 27, 2016

Answer:

#x=-3/2#

Explanation:

Properties of logarithms
#y=log_ax# is the same as #a^y=x#

Your posted equation will be #2x+3=log_4 1#

To solve the right side, if you don't have a calculator that with a change of base: Enter #LOG(1)#/LOG(4) into your calculator to get the correct answer. You divide the log by the number you want to change the base to.

That gives you 2x+3=0. From there it's simple algebra. Subtract the 3 from each side and divide both sides by 2. Leaving you with #x=-3/2# .

Always check your answers. #4^(2*(-3/2)+3# = 1