Precalculus Examples

$\log_{9} (7 x + 5) = \log_{9} (7 x + 2)$

Step 1

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 9$ , $x = 7 x + 5$ , and $y = \log_{9} (7 x + 2)$ .

$b = 9$

$x = 7 x + 5$

$y = \log_{9} (7 x + 2)$

Step 2

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

$9^{\log_{9} (7 x + 2)} = 7 x + 5$