How do I solve $\log_{10}(X^2+3x)=1-log_{10}(x-3)$? Note this is a final exam problem in a basic algebra class.

Question

Basic math takes me to $x^3-9x-10=0$, but what then? Note easy to throw into Wolfram Alpha or something, but how about this in a final?

Answer 1

Hint: $$x^3-9x-10=x^3+2x^2-2x^2-4x-5x-10=\\ x^2(x+2)-2x(x+2)-5(x+2)=\\ (x+2)(x^2-2x-5)$$

Answer 2

Ok. Since you are just trying to learn from an old exam, then I'd happily help you out.

Your first step looks good. Basically, you moved $\log_{10}(x - 3)$ to the left, and got $\log_{10}(x^2 + 3x) + \log_{10}(x - 3) = \log_{10}(x^3 - 9x) = 1$, from which you deduced $x^3 - 9x = 10$. Good.

So the next step is to solve the cubic equation $x^3 - 9x - 10 = 0$. While there is the general Cardano-Tartaglia formula for solving cubics, I doubt that's expected in an algebra class.

In a classroom setting, usually the cubic equations are simple - most likely one of the roots is $0, \pm 1, \pm2, \pm 3$. Just check all of them. For $x^3 - 9x - 10 = 0$, it turns out that $x = -2$ is a root:

$(-2)^3 - 9 \times (-2) - 10 = -8 + 18 - 10 = 0$.

So now we know that $-2$ is a root. That means $x + 2$ is a factor of $x^3 - 9x - 10$. Using, say, long division, you can factorize $x^3 - 9x - 10 = (x + 2) (x^2 - 2x - 5)$.

Now you just need to solve $(x + 2)(x^2 - 2x - 5) = 0$. One root is $x = -2$, as we already found. You get the other roots by solving $x^2 - 2x - 5 = 0$, which are $1 + \sqrt{6}$ and $1 - \sqrt{6}$.

Finally, plug them back into $\log_{10}(x^2 + 3x) = 1 - \log_{10}(x - 3)$ to make sure they make sense. (You need to make sure $x^3 + 3x > 0$ and $x - 3 > 0$. I think that rules out $x = -2$ and $x = 1 - \sqrt{6}$.)