How do you solve #25^ { x ^ { 2} - x } = 1#? Precalculus 1 Answer Monzur R. · Anjali G Mar 25, 2017 #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# Answer link Related questions How do I determine the molecular shape of a molecule? What is the lewis structure for co2? What is the lewis structure for hcn? How is vsepr used to classify molecules? What are the units used for the ideal gas law? How does Charle's law relate to breathing? What is the ideal gas law constant? How do you calculate the ideal gas law constant? How do you find density in the ideal gas law? Does ideal gas law apply to liquids? Impact of this question 254 views around the world You can reuse this answer Creative Commons License