Chapter 3.5, Problem 37E

Solution

To solve: the given equation by two methods.

  $x=4$

Given information:

The equation:

  $\ln \left(x-3\right)+\ln \left(x+4\right)=3\ln 2$

Property Used:

Power rule:

  ${\log }_b{x}^m=m{\log }_{b}x$

Product rule:

  ${\log }_{b}x+{\log }_{b}y={\log }_{b}xy$

One-to-One Properties

For any exponential function $f\left(x\right)=b^x$ ,

If $b^u=b^v$, then $u=v$ .

For any logarithmic function $f\left(x\right)={\log }_{b}x$ ,

If ${\log }_{b}u={\log }_{b}v$, then $u=v$ .

Quadratic Formula:

If $a\ne 0$ the solutions of the equation $a{x}^2+bx+x=0$ are given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Explanation:

First Method:

Consider the given equation:

  $\ln \left(x-3\right)+\ln \left(x+4\right)=3\ln 2$

Now, the logarithm function (for any base) is defined only for positive numbers. So, here for the domain:

  $\begin{array}{c}x-3>0\\ x>3\end{array}$ and $\begin{array}{c}x+4>0\\ x>-4\end{array}$

Domain here is $\left(3,\infty \right)$

Using the power rule, given equation can be rewritten as:

  $\ln \left(x-3\right)+\ln \left(x+4\right)=\ln 2^3$

Using the product rule,

  $\ln \left[\left(x-3\right)\left(x+4\right)\right]=\ln 8$

Using one-to-one property of logarithm, this is equivalent to,

  $\left(x-3\right)\left(x+4\right)=8$

Simplify,

  $\begin{array}{c}{x}^2-3x+4x-12=8\\ x^2+x-20=0\end{array}$

By using quadratic formula, the solutions of this quadratic equation are:

  $\begin{array}{c}x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(-20\right)}}{2\left(1\right)}\\ x=\frac{-1\pm \sqrt{1+80}}{2}\\ x=\frac{-1\pm \sqrt{81}}{2}\end{array}$

  $\begin{array}{c}x=\frac{-1\pm 9}{2}\\ x=-5\text{ or }x=4\end{array}$

Since Domain here is $\left(3,\infty \right)$, so $x\ne -5$

Thus,

  $x=4$

Second method: Graphing both sides of the equation and finding intersections which will be the required solution.