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

1 Answer
Noah G
Feb 24, 2016

Convert to logarithmic form.

Explanation:

#log(4^(2x + 3)) = 1#

Simplify using the rule #logn^a = alogn#

#(2x + 3)log4 = 1#

You can now distribute. Don't forget: you can't combine logarithmic and non logarithmic expressions.

#2xlog4 + 3log4 = 1#

#2xlog4 = 1 - 3log4#

Factor out the x.

#x(2log4) = 1 - log4^3#

#x = (1 - log64)/log16#

Ask your teacher exactly what he/she wants. They may want you to keep your answer in exact form, or for you to round it off to a certain number of decimals, so just make sure to communicate.

Practice exercises:

  1. Solve for x.

a) #2^(3x - 7) = 5#

b). #3^(2x + 3) = 4^(x - 5)#

c) #2^(4x - 9) = 3 xx 3^(x + 6)#

Good luck!

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