How do you solve and find all possible solutions for x?

Question

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

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Answer 1 · 2017-10-01T17:08:05

Either the power must equal 0 or the base must equal 1

Given: #(3x-1)^(x^2-4)=1#

For the given equation to be true, we have these two possible equations:

#(3x-1)^0 = 1# or #(1)^(x^2-4) = 1#

This implies that:

#x^2 - 4 = 0# or #3x-1 = 1#

Here is the next step in solve both equations:

#x^2 = 4# or #3x = 2#

The first equation gives us two values and the second equation, one:

#x = 2, x = -2, or x = 2/3#