How do you solve #25^ { x ^ { 2} - x } = 1#?

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How do you solve #25^ { x ^ { 2} - x } = 1#?

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Answer 1 · 2017-03-25T14:34:13

#x_1=0#

#x_2=1#

Explanation:

Using the law of exponents, we know that the only exponent of #25# which yields an answer of #1# is #0#. #(log_(25)1=0)#

Thus, #x^2-x=0#

#x(x-1)=0#

#x=0#

#x-1=0#

#x=1#

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How do you solve #25^ { x ^ { 2} - x } = 1#?

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