How do you solve $9^{x + 1} = 27$ $9^(x+1) = 27$ ?

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Answer 1 · 2016-04-12T01:48:09

$x = \frac{1}{2}$ $x = 1/2$

We will use the following properties:

$9^{x + 1} = 27$ $9^(x+1) = 27$

$\Rightarrow {log}_{3} (9^{x + 1}) = {log}_{3} (27)$ $=> log_3(9^(x+1)) = log_3(27)$

$\Rightarrow (x + 1) {log}_{3} (9) = {log}_{3} (27)$ $=>(x+1)log_3(9) = log_3(27)$

$\Rightarrow (x + 1) {log}_{3} (3^{2}) = {log}_{3} (3^{3})$ $=>(x+1)log_3(3^2) = log_3(3^3)$

$\Rightarrow 2 (x + 1) = 3$ $=>2(x+1) = 3$

$\Rightarrow x + 1 = \frac{3}{2}$ $=>x+1 = 3/2$

$:. x = 1/2$

Answer 2 · 2016-04-12T01:51:08

Alternative solution:

You can write everything in the same base:

$(3^2)^(x + 1) = 3^3$

$3^(2x + 2) = 3^3$

$3^(2x) = 3^(3 - 2)$

$3^(2x) = 3^1$

$2x = 1$

$x = 1/2$

Hopefully this helps!

How do you solve 9^(x+1) = 279x+1=279^(x+1) = 27?