Question

solve the equation \\(log_{9}(2x - 1) = -1\\). (1 point) \\(\circ\\) \\(x = -\frac{4}{9}\\) \\(\circ\\) \\(x = \frac{5}{9}\\) \\(\circ\\) \\(x = 0\\) \\(\circ\\) \\(x = 5\\)

Explanation:

Step1: Recall the logarithmic definition

If $\log_{a}b = c$, then $b = a^{c}$. Here, $a = 9$, $b = 2x - 1$, and $c = - 1$. So we can rewrite the equation as $2x - 1=9^{-1}$.

Step2: Simplify the right - hand side

We know that $a^{-n}=\frac{1}{a^{n}}$, so $9^{-1}=\frac{1}{9^{1}}=\frac{1}{9}$. Then our equation becomes $2x - 1=\frac{1}{9}$.

Step3: Solve for x

First, add 1 to both sides of the equation: $2x=\frac{1}{9}+1$. We can rewrite 1 as $\frac{9}{9}$, so $\frac{1}{9}+\frac{9}{9}=\frac{1 + 9}{9}=\frac{10}{9}$. Then $2x=\frac{10}{9}$. Next, divide both sides by 2: $x=\frac{10}{9}\div2$. Dividing by 2 is the same as multiplying by $\frac{1}{2}$, so $x=\frac{10}{9}\times\frac{1}{2}=\frac{10\times1}{9\times2}=\frac{5}{9}$.

Answer:

$x = \frac{5}{9}$ (corresponding to the option: $x=\frac{5}{9}$)