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I am asked to simplify $$J=\ln (x^2-16)-\ln x-\ln(x-4), \quad x>4$$ Since each logarithm's argument is non-negative, I can use $$\ln x -\ln y=\ln\frac{x}{y}$$ and obtain the correct answer $$\ln\frac{x+4}{x}$$ However I tried to get the answer by evaluating the rightmost two logarithms first and get $$J=\ln(x^2-16)-\ln\frac{x-4}{x}=\ln\frac{(x^2-16)x}{x-4}$$ which is not the same. What am I doing wrong when evaluating it the second way? Does the order matter?

Jacob
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2 Answers2

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What am I doing wrong when evaluating it the second way?

Note that $$-\ln x-\ln(x-4)=-(\ln x\color{red}{+}\ln(x-4))$$

mathlove
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  • So I've misinterpreted the rule? And $-\ln x-\ln y\neq -\ln \frac{x}{y}$? – Jacob Jun 01 '16 at 15:59
  • @Jacob: $-\ln x-\ln y=-(\ln x\color{red}{+}\ln y)=-\ln(xy)$. – mathlove Jun 01 '16 at 16:00
  • @Jacob The property (implicitly) requires that the first logarithm have coefficient $1$ (not $-1$); $\ln x - \ln y$ is OK, but $-\ln x - \ln y$ needs massaging. – pjs36 Jun 01 '16 at 16:05
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Note that $$\ln x \color{orange}{+} \ln y =\ln xy$$ and $$\ln x \color{blue}{-} \ln y =\ln \frac{x}{y}$$

$$\begin{align} & J=\ln (x^2-16)-\ln x-\ln(x-4) \\ & =\ln (x^2-16)-[\ln x+\ln(x-4)] \\ & =\ln (x^2-16)-[\ln x(x-4)] \\ & =\ln (x^2-16)-\ln x(x-4) \\ & =\ln \frac{(x^2-16)}{x(x-4)} \\ & =\ln \frac{(x-4)(x+4)}{x(x-4)} \\ & =\ln \frac{x+4}{x}\end{align}$$

SchrodingersCat
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