How do you solve #9^ { 3y } = 27^ { 2y + 2}#?

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Answer 1 · 2017-04-19T20:39:39

There is no solution.

Write the bases #9# and #27# as powers of #3#:

#9^(3y)=27^(2y+2)#

So:

#(3^2)^(3y)=(3^3)^(2y+2)#

Use #(a^b)^c=a^(bc)#:

#3^(6y)=3^(6y+6)#

The powers must be equal:

#6y=6y+6#

#0=6#

There is no solution.