Pait 1 of 3 Model the pair of situations with exponential functions \( f \) and \( g \). Find the approximate value of x that makes \( \mathrm{f}(\mathrm{x}) \) \( =\mathrm{g}(\mathrm{x}) \). f initial value of 1,000 decreasing at a rate of \( 15 \% \) g initial value of 20 increasing at a rate of \( 15 \% \) Write the exponential function for f . \( \mathrm{f}(\mathrm{x})=\square \) (Use integers or decimals for any numbers in the expression.)

To model the situation with the exponential function for \( f \), we can use the formula: \[ f(x) = f_0 \cdot (1 - r)^x \] where \( f_0 \) is the initial value and \( r \) is the rate of decrease expressed as a decimal. For \( f \): - Initial value \( f_0 = 1000 \) - Rate of decrease \( r = 0.15 \) Substituting these values into the formula gives: \[ f(x) = 1000 \cdot (1 - 0.15)^x = 1000 \cdot (0.85)^x \] So, we have \( f(x) = 1000 \cdot (0.85)^x \). Now, we can derive the exponential function for \( g \): \[ g(x) = g_0 \cdot (1 + r)^x \] For \( g \): - Initial value \( g_0 = 20 \) - Rate of increase \( r = 0.15 \) Substituting these values gives: \[ g(x) = 20 \cdot (1 + 0.15)^x = 20 \cdot (1.15)^x \] So, we have \( g(x) = 20 \cdot (1.15)^x \). To find the approximate value of \( x \) that makes \( f(x) = g(x) \): \[ 1000 \cdot (0.85)^x = 20 \cdot (1.15)^x \] Dividing both sides by 20: \[ 50 \cdot (0.85)^x = (1.15)^x \] Taking the logarithm of both sides could help us solve for \( x \). However, let’s not jump too far ahead just yet! Grab your calculator, and let's estimate \( x \) where these two functions meet!