Chapter 3.5, Problem 36E

Solution

To solve: the given equation by two methods.

  $x\approx 103.8517$

Given information:

The equation:

  $\log x-\frac{1}{2}\log \left(x+4\right)=1$

Definition Used:

Common logarithm- Base 10:

The common logarithmic function ${\log }_{10}x=\log x$ is the inverse of the exponential function $f\left(x\right)={10}^x$ . Therefore,

  $y=\log x$ if and only if ${10}^y=x$

Property Used:

Quotient rule:

  ${\log }_b\left(\frac{x}{y}\right)={\log }_{b}x-{\log }_{b}y$

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:

  $\log x-\frac{1}{2}\log \left(x+4\right)=1$

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

  $\log x-\log {\left(x+4\right)}^{\frac{1}{2}}=1$

Using quotient rule, this is equivalent to,

  $\log \left(\frac{x}{\sqrt{x+4}}\right)=1$

By definition of common logarithm, this is equivalent to,

  $\frac{x}{\sqrt{x+4}}={10}^1$

Simplify,

  $\begin{array}{c}\frac{x}{\sqrt{x+4}}=10\\ x=10\sqrt{x+4}\end{array}$

Squaring both sides of the above equation,

  $\begin{array}{c}{x}^2={10}^2\left(x+4\right)\\ x^2=100x+400\\ x^2-100x-400=0\end{array}$

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

  $\begin{array}{c}x=\frac{-\left(-100\right)\pm \sqrt{{\left(-100\right)}^2-4\left(1\right)\left(-400\right)}}{2\left(1\right)}\\ x=\frac{100\pm \sqrt{10000+1600}}{2}\\ x=\frac{100\pm \sqrt{11600}}{2}\\ x=\frac{100\pm 20\sqrt{29}}{2}\end{array}$

Since natural logarithm is not defined for negative values, so $x\ne \frac{100-20\sqrt{29}}{2}$

Thus,

  $\begin{array}{l}x=\frac{100+20\sqrt{29}}{2}\\ x\approx 103.8517\end{array}$

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