Home > Precalculus > Logarithmic Models

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

Answer 1

Cora Alexander

For this problem, we must use the rule #a^n = b^m -> loga^n = logb^m#

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

#log4^(2x + 3) = log1#

Use the property #logn^a = alogn#

#(2x + 3)log4 = 0#

#2x + 3 = 0/log4#

#2x + 3 = 0#

#2x = -3#

#x = -3/2#

Checking we find that the solution works, since #a^0 = 1 -> 4^0 = 1#

Practice tasks:

Solve for x. Express your answer in terms of #logx# (exact values)

a) #2^(3x + 1) = 5#

b) #3^(2x - 1) = 4^(3x - 7)#

Challenge Issue

Solve #2^(3x) xx 5 = 3^(4x - 1)#

Wishing you luck!

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Trending questions

Not the question you need?

HIX Tutor

Solve ANY homework problem with a smart AI

98% accuracy study help
Covers math, physics, chemistry, biology, and more
Step-by-step, in-depth guides
Readily available 24/7