Chapter 5, Problem 17CT

To determine

The solution of the equation $\log \left(x^2+3\right)=\log \left(x+6\right)$. Express irrational solutions in exact form.

Answer to Problem 17CT

Solution:

The solution of the equation $\log \left(x^2+3\right)=\log \left(x+6\right)$ is $x=\frac{1+\sqrt{13}}{2}\text{or}x=\frac{1-\sqrt{13}}{2}$.

Explanation of Solution

Given information:

The equation is $\log \left(x^2+3\right)=\log \left(x+6\right)$.

Explanation:

Let $\log \left(x^2+3\right)=\log \left(x+6\right)$.

The domain of the variable requires that $x^2+3>0$ and $x+6>0$, so $\left\{x|x\in ℝ\right\}$ and $x>-6$.

This means, any solution must satisfy all real number, or equivalently, $\left(-\infty ,\infty \right)$ in interval.

It can be written as ${\log }_{10}\left(x^2+3\right)={\log }_{10}\left(x+6\right)$.

By using if ${\log }_{a}M={\log }_{a}N$, then $M=N$ where $M,N,a>0$, and $a\ne 1$,

Here $a=10>0$ and $a\ne 1$.

$\Rightarrow x^2+3=x+6$

Subtracting $x+6$ from both sides,

$\Rightarrow x^2+3-x-6=0$

$\Rightarrow x^2-x-3=0$

To solve this quadratic equation, use quadratic formula.

By using quadratic formula,

$\Rightarrow x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-3\right)}}{2\left(1\right)}$

$\Rightarrow x=\frac{1\pm \sqrt{13}}{2}$

$\Rightarrow x=\frac{1+\sqrt{13}}{2}\text{or}x=\frac{1-\sqrt{13}}{2}$

Thus, the solution of equation $\log \left(x^2+3\right)=\log \left(x+6\right)$ is $x=\frac{1+\sqrt{13}}{2}\text{or}x=\frac{1-\sqrt{13}}{2}$.