How do you solve #25^x=125^(x-2)#?

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Answer 1 · 2017-06-05T07:15:03

#x=6#

In this exponential equation, the clue is that both #25 and 125# are powers of #5#.

Write them both with a base of #5#

#25^x = 125^(x-2)#

#(5^2)^x = (5^3)^(x-2)#

Raising a power to a power #rarr#multiply the indices:

#color(blue)(5)^color(red)((2x)) = color(blue)(5)^color(red)((3x-6))#

In this equation the bases are both equal (to #5#,) so the indices also have to be equal.

#:. color(red)(2x=3x-6)" "larr# solve this linear equation for #x#

#6=x#