Chapter 3, Problem 53RE

Solution

To solve: the given equation.

  $x\approx 99.5112$

Given information:

The equation: $\log \left(x+2\right)+\log \left(x-1\right)=4$

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:

Product rule:

  ${\log }_b\left(x\cdot y\right)={\log }_{b}x+{\log }_{b}y$

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:

Consider the given equation:

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

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

  $\log \left[\left(x+2\right)\left(x-1\right)\right]=4$

By the definition of common logarithm, this is equivalent to,

  $\left(x+2\right)\left(x-1\right)={10}^4$

Simplify,

  $\begin{array}{c}{x}^2-x+2x-2={10}^4\\ x^2+x-2=10000\\ x^2+x-10002=0\end{array}$

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

  $\begin{array}{c}x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(-10002\right)}}{2\left(1\right)}\\ =\frac{-1\pm \sqrt{1+40008}}{2}\\ =\frac{-1\pm \sqrt{40009}}{2}\end{array}$

Now, as logarithm is not defined for negative values, so $x>0$

Therefore,

  $x\ne \frac{-1-\sqrt{40009}}{2}$

Hence, solution is:

  $\begin{array}{c}x=\frac{-1+\sqrt{40009}}{2}\\ x\approx 99.5112\end{array}$