Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Test - Page 496: 16

$0$ and $ln(\frac{3}{2})\approx0.405$

Step 1. Let $u=e^x$, we have $2u^2-5u+3=0\longrightarrow (u-1)(2u-3)=0$, thus $u=1, \frac{3}{2}$ Step 2. For $u=e^x=1$, we have $x=ln1=0$ Step 3. For $u=e^x=\frac{3}{2}$, we have $x=ln(\frac{3}{2})=ln(1.5)\approx0.405$