Dear Uncle Colin,

I’m trying to solve $3^{2x+1} - 28\times 3^x + 9 =0$. I’ve split up the first term into $3 \times 3^{2x}$ but then I’m stuck! Any suggestions?

Likely Overthinking Gettable Sum

Hi, LOGS, and thanks for your message!

You’ve made a really good start there. The piece you’re missing is to note that $3^{2x}$ is $\left(3^x\right)^2$.

How does that help?

It helps because then you have a quadratic. I find it’s simplest to make $y=3^x$ in this sort of situation, because then you can write the equation as $3y^2 - 28y + 9 = 0$.

That factorises as $(3y-1)(y-9)=0$, so $y=\frac{1}{3}$ or $y=9$.

But we made up $y$!

So we did. We’re not finished until we turn back to our original variable, $3^x = \frac{1}{3}$ or $3^x = 9$; this gives us $x=-1$ or $x=2$.

Hope that helps,

- Uncle Colin