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

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Answer 1 · 2016-08-15T02:46:12

#ln(4^(3x + 1)) = ln1#

#(3x + 1)ln4 = ln1#

#3x + 1 = ln1/ln4#

#3x + 1 = 0/ln4#

#3x + 1 = 0#

#3x = -1#

#x = -1/3#

Checking in the original equation, we find that #x = -1/3# works.

Hopefully this helps!