How do you solve #9^(x-1)times81^(2x-1)=27^(3x-2)#?

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Answer 1 · 2016-11-30T16:04:23

#x = 0#

Write in the same bases.

#(3^2)^(x- 1) xx (3^4)^(2x - 1) = (3^3)^(3x - 2)#

#3^(2x- 2) xx 3^(8x- 4) = 3^(9x - 6)#

You can now use the multiplication rule of exponents that #a^m xx a^n = a^(m + n)#.

#3^(2x - 2 + 8x - 4) = 3^(9x - 6)#

We are in equivalent bases so we can eliminate.

#2x - 2 + 8x- 4= 9x - 6#

#x = 0#

Hopefully this helps!