Model the pair of situations with exponential functions \( f \) and \( g \). Find the approximate value of \( x \) tha \( =g(x) \). f . initial value of 700 decreasing at a rate of \( 15 \% \) \( g \) : initial value of 20 increasing at a rate of \( 15 \% \) wrme me exponenual runction ror \( g \). \( g(x)=20(1.15)^{x} \) (Use integers or decimals for any numbers in the expression.) The value of \( x \) that makes \( f(x)=g(x) \) is \( x=\square \). (Round to the nearest integer as needed.)

To model the pair of situations with exponential functions, we start by developing the equations: For \( f(x) \), which is the initial value of 700 decreasing at a rate of 15%, the function can be modeled as: \[ f(x) = 700(0.85)^{x} \] For \( g(x) \), which is the initial value of 20 increasing at a rate of 15%, you already have: \[ g(x) = 20(1.15)^{x} \] Now, we need to find the value of \( x \) that makes \( f(x) = g(x) \): \[ 700(0.85)^{x} = 20(1.15)^{x} \] To solve for \( x \), we can rearrange and simplify: \[ \frac{700}{20} = \frac{(1.15)^{x}}{(0.85)^{x}} \] \[ 35 = \left(\frac{1.15}{0.85}\right)^{x} \] \[ 35 = \left(\frac{1.15}{0.85}\right)^{x} \] Now, taking the logarithm of both sides: \[ \log(35) = x \cdot \log\left(\frac{1.15}{0.85}\right) \] \[ x = \frac{\log(35)}{\log\left(\frac{1.15}{0.85}\right)} \] Calculating the right-hand side gives approximately: \[ x \approx 17.056 \] Rounding to the nearest integer: The value of \( x \) that makes \( f(x) = g(x) \) is \( x = 17 \).