Answer
$h^{-1}(x) = \frac{\ln (\frac{x}{-3.5})}{\ln 1.8}$
Work Step by Step
$h(x) = -3.5(1.8)^{x}$
Let $h(x) = y$
$y = -3.5(1.8)^{x}$
Swap the variables $x$ and $y$ and then solve for $y$ to find the inverse:
$x = -3.5(1.8)^{y}$
$(1.8)^{y} = -\frac{x}{3.5}$
$y = \ln_{1.8} (\frac{x}{-3.5})$
$y = \frac{\ln (\frac{x}{-3.5})}{\ln 1.8}$
$h^{-1}(x) = \frac{\ln (\frac{x}{-3.5})}{\ln 1.8}$
