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

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Answer 1 · 2016-10-24T02:18:09

Put in a common base.

$3^(5x) xx (3^4)^(1 - x) = (3^2)^(x - 3)$

Simplify using the rules $a^n xx a^m = a^(n + m)$ and $(a^n)^m = a^(n xx m)$ .

$3^(5x) xx 3^(4 - 4x) = 3^(2x - 6)$

$3^(5x + 4 - 4x) = 3^(2x - 6)$

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

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

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

$-x = -10$

$x = 10$

Hopefully this helps!

Answer 2 · 2016-10-24T02:23:37

$x = 10$

$3^(5x)*81^(1-x) = 9^(x-3)$

$=> 3^(5x)*(3^4)^(1 - x) = (3^2)^(x-3)$

$=> 3^(5x)*3^(4(1-x)) = 3^(2(x-3))$

$=>3^(5x + 4 - 4x) = 3^(2x - 6)$

$=> 3^(x + 4) = 3^(2x - 6)$

Apply $log_3$ to both sides

$=> log_3 3^(x + 4) = log_3 3^(2x - 6)$

$=> (x + 4) log_3 3 = (2x - 6)log_3 3$

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

$=> x = 10$

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