Dear Uncle Colin,

How would you solve $5^{2x+2} + 16\cdot15^x -9^{x+1} = 0$?

Doesn’t It Seem Genuinely Unpleasant? Insoluble? Strange, Even?

Hi, Disguise, and thanks for your message! It turns out that this does have solutions!

The trick is to ask “what would make it less unpleasant?” - and given that all of the base numbers are multiples of powers of 3, 5 or both, it seems worth saying "Let $A = 5^x$ and $B = 3^x$" and seeing what drops out.

$5^{2x+2} = 25 \cdot 5^{2x}$, or $25A^2$.

$16\cdot 15^x = 16 \cdot AB$

$-9^{x+1} = -9B^2$.

So the equation can be written as $25A^2 + 16AB - 9B^2=0$.

It may not be obvious, but that can be factorised as $(25A-9B)(A+B)=0$.

So either $A = -B$ - however, $A$ and $B$ are both positive, so this can’t be true - or $25A = 9B$.

Rewriting that, $5^{x+2} = 3^{x+2}$.

Taking logs: $(x+2)\ln(5) = (x+2)\ln(3)$.

Now, $\ln(5)$ certainly doesn’t equal $\ln(3)$, so $x+2$ must be zero - and $x = -2$ is the only solution to this.

Checking, $5^{-2} + 16\cdot 15^{-2} - 9^{-1}$ is $\frac{1}{25} + \frac{16}{225} - \frac{1}{9}$.

That’s $\frac{9}{225} + \frac{16}{225} - \frac{25}{225} = 0$, so it does indeed hold.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.