Precalculus Examples

$(\log_{7} (x + 3) \div \log_{7} (x + 1)) = (\log_{7} (x - 2 \div \log_{7} (x + 4)))$

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, $b = 7$ , $x =$ , and $y = \log_{7} (x - 2 \div \log_{7} (x + 4))$ .

$b = 7$

$x =$

$y = \log_{7} (x - 2 \div \log_{7} (x + 4))$

Step 2

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

$7^{\log_{7} (x - 2 \div \log_{7} (x + 4))} =$