Chapter 6 - Logarithmic Functions - 6.2 Logarithmic Functions - 6.2 Exercises - Page 504: 95

$h^{-1}(x) = \log_9 2x$

$h(x) = \frac{1}{2}(9^{x})$ Let $h(x) = y$ $y = \frac{1}{2}(9^{x})$ Swap the $x$ and $y$ variable to find the inverse: $x = \frac{1}{2}(9^{y})$ $x = \frac{9^{y}}{2}$ $2x = 9^{y}$ $y = \log_9 2x$ $h^{-1}(x) = \log_9 2x$