How do you solve #125^(2x+1)/625^(x+2)=3125^(x+2)#?

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Answer 1 · 2016-09-19T06:33:32

#x=-5#

#(125^(2x+1))/(625^(x+2))=3125^(x+2)#

or #((5^3)^(2x+1))/((5^4)^(x+2))=(5^5)^(x+2)#

or #(5^(3xx(2x+1)))/(5^(4xx(x+2)))=5^(5xx(x+2)#^

or #(5^(6x+3))/(5^(4x+8))=5^(5x+10)#

or #5^(6x+3-4x-8)=5^(5x+10)#

or #5^(2x-5)=5^(5x+10)#

or #2x-5=5x+10#

or #2x-5x=10+5#

or #-3x=15#

or #x=15/-3=-5#