For the problem below, I only need help on expalnation for part b.
For the first problem, graph the 2 equations. It is then easy to see that there are 2 solutions: one negative and one positive.
For the second problem, take logarithms on both sides
$\displaystyle (x+2)ln 3=(2x-1)ln 5$
$\displaystyle \frac{x+2}{2x-1}=\frac{ln 5}{ln 3}$
For part b),
i) $\displaystyle (f+g)(0)=f(0)+g(0)=2+1=3$
ii) $\displaystyle (fg)(0)=f(0)g(0)=2*1=2$