How do you solve # 9^(2x)=27^(x-1)#?

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Answer 1 · 2015-08-19T06:10:42

#color(blue)(x=-3#

#9^(2x)=27^(x−1#

We know that
#9=3^2#
#27=3^3#

So,

#3^(2(2x))=3^(3(x−1)#

#3^(4x)=3^(3x−3)#

Now as bases are equal we can equate exponents

#4x=3x-3#
#color(blue)(x=-3#