Setting the two equations equal yields $3^{2x}-3^x = 4*3^x$
Let $y=3^x$
Then we have $y^2-y=4y$
$y(y-5)=0$
$y=0,5$
$3^x = 0, 3^x =5$
$x\log 3 =0, \implies x=0$ and $x\log 3 = 5 \implies x = \frac{5}{\log 3}$
Is this correct?
How can I tell which graph is which just by looking at it?
Doing the same thing for the next question gives
$y^2-y-20 = (y-5)(y+4) \implies y = 5, y=4$
$3^x = 5,$ and $ 3^x = 4$ so
$x\log 3 = 5 \implies x = \frac{5}{\log 3}$ and $x\log 3 = 4 \implies x = \frac{4}{\log 3}$