Chapter 7 Exponential and Logarithmic Functions - 7.6 Solve Exponential and Logarithmic Equations - 7.6 Exercises - Problem Solving - Page 522: 60a

2.7999 cm

$I(x)=I_0e^{-\mu x}$ Solve for x: $\ln I=\ln(I_0e^{-\mu x})=\ln I_0+\ln e^{-\mu x}\\ \rightarrow -\mu x=\ln I-\ln I_0\\ \rightarrow x=-\frac{\ln\frac{I}{I_0}}{\mu}$ Substitute $I=0.3I_0\\\mu=0.43$ We have: $x=-\frac{\ln\frac{0.3I_0}{I_0}}{0.43}=-\frac{-\ln 0.3}{0.43}\approx 2.7999$