3
$\begingroup$

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?

  • 1
    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).$2011-11-29

2 Answers 2

4

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 $x^2-16$?

  • 1
    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)$2011-11-29
2

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.