Chapter 3.5, Problem 35E

Solution

To solve: the given equation by two methods.

  $x\approx 2.3028$

Given information:

The equation:

  $\frac{1}{2}\ln \left(x+3\right)-\ln x=0$

Property Used:

Power rule:

  ${\log }_b{x}^m=m{\log }_{b}x$

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:

  $\begin{array}{c}\frac{1}{2}\ln \left(x+3\right)-\ln x=0\\ \frac{1}{2}\ln \left(x+3\right)=\ln x\\ \ln \left(x+3\right)=2\ln x\end{array}$

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

  $\ln \left(x+3\right)=\ln x^2$

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

  $x+3=x^2$

Simplify,

  $x^2-x-3=0$

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

  $\begin{array}{c}x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-3\right)}}{2\left(1\right)}\\ x=\frac{1\pm \sqrt{1+12}}{2}\\ x=\frac{1\pm \sqrt{13}}{2}\end{array}$

Since natural logarithm is not defined for negative values, so $x\ne \frac{1-\sqrt{13}}{2}$

Thus,

  $\begin{array}{l}x=\frac{1+\sqrt{13}}{2}\\ x\approx 2.3028\end{array}$

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