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

Question

4149 views around the world

You can reuse this answer
Creative Commons License

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!

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