Given \( h(x)=\left(\frac{1}{s}\right)^{x} \) 7.3.1 Determine the equation of \( x-1(x) \), the inverse of \( h \), in the forn \( y=\ldots \) 7.3.2 Describe the transformation from, 8 to \( h \). i

We are given the function   h(x) = (1/s)^x. Below are the steps to solve each part. ────────────────────────────── 1. Inverse Function of h(x) To find the inverse function, h⁻¹(x), we start with   y = (1/s)^x. Step 1. Interchange x and y (since the inverse “swaps” the roles of the input and output):   x = (1/s)^y. Step 2. Solve for y. To do so, take the natural logarithm on both sides:   ln(x) = ln((1/s)^y). Step 3. Use the logarithm power rule (ln(a^b) = b·ln(a)):   ln(x) = y · ln(1/s). Step 4. Solve for y:   y = ln(x) / ln(1/s). Thus, one acceptable form for the inverse is   h⁻¹(x) = ln(x) / ln(1/s). An equivalent form can be obtained by recalling that ln(1/s) = –ln s. Therefore, we may also write   h⁻¹(x) = –ln(x)/ln(s). Yet another equivalent way to express the answer is to use logarithms with base s:   h⁻¹(x) = –logₛ x. ────────────────────────────── 2. Transformation from the Base Exponential to h(x) Observe that   h(x) = (1/s)^x = s^(–x). This means that h(x) is obtained from the usual exponential function f(x)=s^x by replacing x with –x. Geometrically, replacing x with –x reflects the graph about the y-axis. In summary, the transformation from the graph of f(x)=s^x to the graph of h(x)=s^(–x) is:   • A reflection across the y–axis. ────────────────────────────── Final Answers 7.3.1 The inverse of h(x) is   h⁻¹(x) = ln(x) / ln(1/s)    or  h⁻¹(x) = –ln(x)/ln(s)  or  h⁻¹(x) = –logₛ x. 7.3.2 The function h(x) = (1/s)^x is obtained from f(x)=s^x by reflecting its graph in the y-axis (i.e. replacing x with –x).