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

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Answer 1 · 2016-10-24T02:18:09

Put in a common base.

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

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

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

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

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

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

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

$- x = - 10$ $-x = -10$

$x = 10$ $x = 10$

Hopefully this helps!

Answer 2 · 2016-10-24T02:23:37

$x = 10$ $x = 10$

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

$\Rightarrow 3^{5 x} \cdot {(3^{4})}^{1 - x} = {(3^{2})}^{x - 3}$ $=> 3^(5x)*(3^4)^(1 - x) = (3^2)^(x-3)$

$\Rightarrow 3^{5 x} \cdot 3^{4 (1 - x)} = 3^{2 (x - 3)}$ $=> 3^(5x)*3^(4(1-x)) = 3^(2(x-3))$

$\Rightarrow 3^{5 x + 4 - 4 x} = 3^{2 x - 6}$ $=>3^(5x + 4 - 4x) = 3^(2x - 6)$

$\Rightarrow 3^{x + 4} = 3^{2 x - 6}$ $=> 3^(x + 4) = 3^(2x - 6)$

Apply ${log}_{3}$ $log_3$ to both sides

$\Rightarrow {log}_{3} 3^{x + 4} = {log}_{3} 3^{2 x - 6}$ $=> log_3 3^(x + 4) = log_3 3^(2x - 6)$

$\Rightarrow (x + 4) {log}_{3} 3 = (2 x - 6) {log}_{3} 3$ $=> (x + 4) log_3 3 = (2x - 6)log_3 3$

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

$\Rightarrow x = 10$ $=> x = 10$

How do you solve 35x⋅811−x=9x−33^(5x)*81^(1-x)=9^(x-3)?