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

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Answer 1 · 2016-04-02T01:25:05

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 exercises:

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 Problem

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

Good luck!