Chapter 6.6, Problem 32E

To determine

To calculate: .The equation for the value of $x$ .

Answer to Problem 32E

The required values of $x$ are $-3$ and $-6$ .

Explanation of Solution

Given Information:

The given equation is ${\log }_3\left(x^2+9x+27\right)=2$ .

Formula Used:

Property of logarithms:

If ${\log }_a\left(x\right)=b$ then $x=a^b$

Quadratic formula:

If $a{x}^2+bx+c=0$ then, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Where $a,b,c$ are constant and $a\ne 0$ .

Calculation:

Consider the equation ${\log }_3\left(x^2+9x+27\right)=2$ .

Use the property of logarithms of the above equation.

  $x^2+9x+27=3^2$

Simplify the equation.

  $\begin{array}{c}{x}^2+9x+27=3^2\\ x^2+9x+27-9=0\\ x^2+9x+18=0\end{array}$

Now, use quadratic formula in the above equation.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $9$ for $b$ and $18$ for $c$ in the above quadratic formula.

  $x=\frac{-9\pm \sqrt{{\left(9\right)}^2-4\left(1\right)\left(18\right)}}{2\left(1\right)}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-9\pm \sqrt{{\left(9\right)}^2-4\left(1\right)\left(18\right)}}{2\left(1\right)}\\ =\frac{-9\pm \sqrt{81-72}}{2}\\ =\frac{-9\pm \sqrt{9}}{2}\\ =\frac{-9\pm \sqrt{3^2}}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-9\pm \sqrt{3^2}}{2}\\ =\frac{-9\pm 3}{2}\\ =\frac{-9+3}{2},\frac{-9-3}{2}\\ =-3,-6\end{array}$

Hence, the required values of $x$ are $-3$ and $-6$ .