# How do you solve 3^(5x)*81^(1-x)=9^(x-3)?

Oct 24, 2016

Put in a common base.

${3}^{5 x} \times {\left({3}^{4}\right)}^{1 - x} = {\left({3}^{2}\right)}^{x - 3}$

Simplify using the rules ${a}^{n} \times {a}^{m} = {a}^{n + m}$ and ${\left({a}^{n}\right)}^{m} = {a}^{n \times m}$.

${3}^{5 x} \times {3}^{4 - 4 x} = {3}^{2 x - 6}$

${3}^{5 x + 4 - 4 x} = {3}^{2 x - 6}$

We can now eliminate the bases and solve like a linear equation.

$5 x + 4 - 4 x = 2 x - 6$

$5 x - 4 x - 2 x = - 6 - 4$

$- x = - 10$

$x = 10$

Hopefully this helps!

Oct 24, 2016

$x = 10$

#### Explanation:

${3}^{5 x} \cdot {81}^{1 - x} = {9}^{x - 3}$

$\implies {3}^{5 x} \cdot {\left({3}^{4}\right)}^{1 - x} = {\left({3}^{2}\right)}^{x - 3}$

$\implies {3}^{5 x} \cdot {3}^{4 \left(1 - x\right)} = {3}^{2 \left(x - 3\right)}$

$\implies {3}^{5 x + 4 - 4 x} = {3}^{2 x - 6}$

$\implies {3}^{x + 4} = {3}^{2 x - 6}$

Apply ${\log}_{3}$ to both sides

$\implies {\log}_{3} {3}^{x + 4} = {\log}_{3} {3}^{2 x - 6}$

$\implies \left(x + 4\right) {\log}_{3} 3 = \left(2 x - 6\right) {\log}_{3} 3$

$\implies x + 4 = 2 x - 6$

$\implies x = 10$