Precalculus Examples

$\log_{16} (\sqrt{x + 1}) = (\frac{1}{4})$

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, $b = 16$ , $x = \sqrt{x + 1}$ , and $y = \frac{1}{4}$ .

$b = 16$

$x = \sqrt{x + 1}$

$y = \frac{1}{4}$

Step 2

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

$16^{\frac{1}{4}} = \sqrt{x + 1}$