How do you solve #25^ { 1- 3n } = 625^ { - n - 1}#?

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Answer 1 · 2017-09-22T05:05:53

#n=3#

Rewrite the expression by finding the same base on both sides

#25^(1-3n)=625^(-n-1)#

#rArr5^(2(1-3n))=5^(4(-n-1))#

#rArr5^(2-6n)=5^(-4n-4)#

Equate the powers to solve for #n#

#2-6n=-4n-4#

#2n=6#

#n=3#