Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 404: 46

$x=2$ or $x=-1$

We need to solve: $x^{2}10^{x}-x10^{x}=2(10^{x})$ $x^{2}10^{x}-x10^{x}-2(10^{x})=0$ We factor: $10^{x}(x^{2}-x-2)=0$ $10^{x}(x-2)(x+1)=0$ $10^{x}=0$ or $x-2=0$ or $x+1=0$ $x=\log_{10} (0)$=no solution, or $x=2$ or $x=-1$