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

Nov 30, 2016

$x = 0$

Explanation:

Write in the same bases.

${\left({3}^{2}\right)}^{x - 1} \times {\left({3}^{4}\right)}^{2 x - 1} = {\left({3}^{3}\right)}^{3 x - 2}$

${3}^{2 x - 2} \times {3}^{8 x - 4} = {3}^{9 x - 6}$

You can now use the multiplication rule of exponents that ${a}^{m} \times {a}^{n} = {a}^{m + n}$.

${3}^{2 x - 2 + 8 x - 4} = {3}^{9 x - 6}$

We are in equivalent bases so we can eliminate.

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

$x = 0$

Hopefully this helps!