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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?

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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 $x\log 3 = 4 \implies x = \frac{4}{\log 3}$

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    For the first solution, the condition $3^x=0$ is not satisfied by any real number (since $\ln(0)$ is undefined) hence that equation yields no solution. $3^x=5\implies \ln(3^x)=\ln(5)\implies x\ln(3)=\ln(5)\implies x=\frac{\ln(5)}{\ln(3)}$. The same method applies to solving the equation $3^x=4$. – coreyman317 May 12 '19 at 04:22
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    ah thank you! i forgot to take the log on both sides. and i forgot the about ln(0) being undefined – user130306 May 12 '19 at 04:23
  • Because $\ 3^{2x}-3^x=3^x(3^x-1)> 4\cdot3^x\ $ whenever $\ x\ $ is sufficiently large ($\ x\ge2\ $ will do), this tells you that the graph that's higher for large values of $\ x\ $ (i.e. the red one) is $\ f(x)$'s. – lonza leggiera May 12 '19 at 05:44
  • @lonzaleggiera but when you choose $x\geq 2$ the blue graph is higher than the red one – user130306 May 12 '19 at 05:58
  • @user130306. No, the point of intersection of the two graphs is $\ x=\frac{\log 5}{\log 3}\approx 1.46< 2\ $. For all points to the right of that point of intersection the red graph is higher than the blue one. – lonza leggiera May 12 '19 at 06:17
  • so for questions like this, will i always have to look past the points of intersection? – user130306 May 12 '19 at 06:18
  • Not necessarily. To recognise which graph belongs to which function just by looking at them, you just have to pick some distinguishing feature of the graphs that is easy to check just by looking at their equations. You could simply look at some particular value of $\ x\ $ for which $\ f(x)\ $ and $\ g(x)\ $ are easy to evaluate just by looking at their equations. Often, as In this case, $\ x=0\ $ will do. Since $\ f(0) =0\ $ and $\ g(0)=4\ne 0\ $ this tells you that the graph of $\ f\ $ is the one passing through the origin—namely the red one. – lonza leggiera May 12 '19 at 06:38

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