Question

what is the exponential function for which the common logarithmic function $h(x) = -\log(3x + 4)$ is the inverse?
$\bigcirc \\ h^{-1}(x) = \frac{1}{3}(10^{-x} - 4)$
$\bigcirc \\ h^{-1}(x) = \frac{1}{310^x} - 4$
$\bigcirc \\ h^{-1}(x) = \frac{1}{10^x}$
$\bigcirc \\ h^{-1}(x) = \frac{1}{2}(10^x - 4)$

Explanation:

Step1: Set $y = h(x)$

$y = -\log(3x + 4)$

Step2: Swap $x$ and $y$

$x = -\log(3y + 4)$

Step3: Isolate the log term

$-x = \log(3y + 4)$

Step4: Convert to exponential form

$10^{-x} = 3y + 4$

Step5: Solve for $y$

$3y = 10^{-x} - 4$
$y = \frac{1}{3}(10^{-x} - 4)$

Answer:

$h^{-1}(x) = \frac{1}{3}(10^{-x} - 4)$ (the first option)