Precalculus Examples

$- 26.99 = 10 \log (\frac{x}{5 {(10^{- 6})}^{2}})$

Divide both sides by $10$ to remove it from the front of the logarithm.

$- \frac{26.99}{10} = \frac{10 \log (\frac{x}{5 {(10^{- 6})}^{2}})}{10}$

Simplify each side of the equation.

$- 2.699 = \log (\frac{x}{5 {(10^{- 6})}^{2}})$

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, $b = 10$ , $x = \frac{x}{5 {(10^{- 6})}^{2}}$ , and $y = - 2.699$ .

$b = 10$

$x = \frac{x}{5 {(10^{- 6})}^{2}}$

$y = - 2.699$

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

$10^{- 2.699} = \frac{x}{5 {(10^{- 6})}^{2}}$