# Expressing as a single logarithm

I've got the equation:

$$\log_{10}(x^2 - 16) - 3\log_{10}(x + 4) + 2\log_{10} x$$

I'm looking to express this as a single logarithm. I came up with

$$\log_{10}(x^2 - 16) - \log_{10}(x + 4)^3 + \log_{10} x^2$$

then

$$\log_{10} \left(\frac{x^2(x^2 - 16)}{(x + 4)^3}\right)$$
Please forgive me if I got the number of parentheses wrong.

This looks like the results of most of the examples, would you think further simplification is required?

• Looks fine to me. Your instructor might wish you to observe that $x^2-16=(x-4)(x+4)$, which gives $\log_{10}\left(\frac{x^2(x-4)}{(x+4)^2}\right).$ Commented Nov 29, 2011 at 1:29

Your reasoning is absolutely right! However, there is one final simplification you can make: to the fraction $$\frac{x^2(x^2 - 16)}{(x + 4)^3}$$ itself. Hint: Can you factor $x^2-16$?

• Would it be correct to then say (x^2(x + 4)(x - 4))/(x + 4)^3 Commented Nov 29, 2011 at 1:30
• Yup! And after cancelling an $x+4$ from numerator and denominator, we get $$\log_{10} \left(\frac{x^2(x - 4)}{(x + 4)^2}\right)$$ Commented Nov 29, 2011 at 1:32

I'm not sure what you mean when you ask if further simplification is required, but it is possible. Your computations are correct; if you notice that $$(x^2 - 16) = (x+4)(x-4),$$ then you can simplify your last expression, $$\log_{10}\left(\frac{x^2(x^2 - 16)}{(x + 4)^3}\right),$$ to $$\log_{10}\left(\frac{x^2(x - 4)}{(x + 4)^2}\right).$$ Not a major simplification, though.

The given answers and hints suggest to cancel out the $$x+4$$ term that is

$$\log_{10}(x^2 - 16) - 3\log_{10}(x + 4) + 2\log_{10} x=\log_{10} \left(\frac{x^2(x^2 - 16)}{(x + 4)^3}\right)=\log_{10} \left(\frac{x^2(x - 4)}{(x + 4)^2}\right)$$

and this is fine.

But note that the equivalence works in this particular case because all the three expressions are defined for $$x>4$$, that is all three have the same domain.

This is not always the case, as for example

$$2\log_{10} x$$

defined for $$x>0$$ is not equivalent to

$$\log_{10} (x^2)$$

which is defined for any $$x\neq 0$$, for that reason we need to pay attention to the domain when we establish an equivalence between different expressions.

For instance, in the previous example we can say that $$\log_{10} (x^2)$$ with $$x>0$$ is equivalent to $$2\log_{10} x$$.