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?
$\endgroup$ 12 Answers
$\begingroup$What am I doing wrong when evaluating it the second way?
Note that $$-\ln x-\ln(x-4)=-(\ln x\color{red}{+}\ln(x-4))$$
$\endgroup$ 3 $\begingroup$Note that $$\ln x \color{orange}{+} \ln y =\ln xy$$ and $$\ln x \color{blue}{-} \ln y =\ln \frac{x}{y}$$
$$\begin{align} & J=\ln (x^2-16)-\ln x-\ln(x-4) \\ & =\ln (x^2-16)-[\ln x+\ln(x-4)] \\ & =\ln (x^2-16)-[\ln x(x-4)] \\ & =\ln (x^2-16)-\ln x(x-4) \\ & =\ln \frac{(x^2-16)}{x(x-4)} \\ & =\ln \frac{(x-4)(x+4)}{x(x-4)} \\ & =\ln \frac{x+4}{x}\end{align}$$
$\endgroup$
