Question

Write each expression as a single logarithm.

$\log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x-2}\right)$

Short answer

The obtained result is$\log \left(\frac{\left(x-1\right)\left(x+3\right)}{\left(x-2\right)\left(x+1\right)\left(x+6\right)}\right)$.

Step-by-step solution

Step 1. Applying the property of logarithm.

Given logarithmic function is $\log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)$ localid="1646898820163" $\log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)$.

Apply the logarithmic property ${\log }_a\left(\frac{M}{N}\right)={\log }_a\left(M\right)-{\log }_a\left(N\right)$.

Here, $M=\frac{x^2+2x-3}{x^2-4}$localid="1646898825116" $M=\frac{x^2+2x-3}{x^2-4}$and $N=\frac{x^2+7x+6}{x+2}$.localid="1646898830194" $N=\frac{x^2+7x+6}{x+2}$.

So, the given function can be written as,

$$\begin{gathered} \log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)=\log \left(\frac{\left(\frac{x^2+2x-3}{x^2-4}\right)}{\left(\frac{x^2+7x+6}{x+2}\right)}\right) \\ =\log \left(\frac{x^2+2x-3}{x^2-4}\times \frac{x+2}{x^2+7x+6}\right) \\ =\log \left(\frac{\left(x^2+2x-3\right)\left(x+2\right)}{\left(x^2-4\right)\left(x^2+7x+6\right)}\right) \\ \log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)=\log \left(\frac{\left(x^2+2x-3\right)\left(x+2\right)}{\left(x^2-4\right)\left(x^2+7x+6\right)}\right)....\left(1\right) \end{gathered}$$

And,

localid="1647003285276" $\begin{array}{rcl}\log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)& =& \log \left(\frac{\left(\frac{x^2+2x-3}{x^2-4}\right)}{\left(\frac{x^2+7x+6}{x+2}\right)}\right)\\ & =& \log \left(\left(\frac{x^2+2x-3}{x^2-4}\right)\times \left(\frac{x+2}{x^2+7x+6}\right)\right)\\ & =& \log \left(\frac{\left(x^2+2x-3\right)\left(x+2\right)}{\left(x^2-4\right)\left(x^2+7x+6\right)}\right)\\ & =& \log \left(\frac{\left(x^2+2x-3\right)\left(x+2\right)}{\left(x^2-4\right)\left(x^2+7x+6\right)}\right)....\left(1\right)\end{array}$

Step 2. Factorizing the terms.

The quadratic $\left(x^2-4\right)$can be factorized as, $(x^2-4)=\left(x-2\right)\left(x+2\right)$.

Also, $\left(x^2+2x-3\right)=\left(x-1\right)\left(x+3\right)$and $\left(x^2+7x+6\right)=\left(x+1\right)\left(x+6\right)$.

Substitute the factors in (1) and simplify.

localid="1646847771926" $$\begin{gathered} \log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)=\log \left(\frac{\left(x-1\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\times \frac{\left(x+2\right)}{\left(x+1\right)\left(x+6\right)}\right) \\ =\log \left(\frac{\left(x-1\right)\left(x+3\right)}{\left(x-2\right)\left(x+1\right)\left(x+6\right)}\right) \end{gathered}$$

Hence the solution is,

localid="1647003294642" $\log \left(\frac{x^2+2x-3}{x^2-4}\right)-\log \left(\frac{x^2+7x+6}{x+2}\right)=\log \left(\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x-2\right)\left(x+6\right)}\right)$