Chapter 3, Problem 112RE

To determine

To solve the logarithmic equation

  ${\log }_5\left(x+2\right)-{\log }_{5}x={\log }_5\left(x+5\right)$

Answer to Problem 112RE

  $x\simeq -4.449\text{ and }x\simeq 0.449$

Explanation of Solution

Given:

  ${\log }_5\left(x+2\right)-{\log }_{5}x={\log }_5\left(x+5\right)$

Concept Used:

For $a\in R^{+}\&a\ne 1$ and $u\in R^{+}$ :

Quotient property of logarithmic function: ${\log }_{a}u-{\log }_{a}v={\log }_a\frac{u}{v}$

  ${\log }_{a}u=x\iff u=a^x$

  $a^{{\log }_{a}x}=x$

Calculation:

In order to solve the logarithmic equation

  ${\log }_5\left(x+2\right)-{\log }_{5}x={\log }_5\left(x+5\right)$

First simplify it using quotient property of logarithmic functions. As

  $\begin{array}{c}{\log }_5\left(x+2\right)-{\log }_{5}x={\log }_5\left(x+5\right)\\ {\log }_5\left(\frac{x+2}{x}\right)={\log }_5\left(x+5\right)\end{array}$

To simplify convert it into exponential equation using the property ${\log }_{a}u=x\iff u=a^x$ .

As,

  $\begin{array}{c}{\log }_5\left(\frac{x+2}{x}\right)={\log }_5\left(x+5\right)\\ \left(\frac{x+2}{x}\right)=e^{{\log }_5\left(x+5\right)}\\ \frac{x+2}{x}=x+5;\text{ since }{a}^{{\log }_{a}x}=x\\ x+2=x\left(x+5\right)\\ x+2=x^2+5x\\ x^2+4x-2=0\\ x=\frac{-4\pm \sqrt{16+4\times 2\times 1}}{2}\\ x=\frac{-4\pm \sqrt{24}}{2}\\ x=-2\pm \sqrt{6}\\ \Rightarrow x\simeq -4.449\text{ and }x\simeq 0.449\end{array}$

Thus, solutions of given equation are $x\simeq -4.449\text{ and }x\simeq 0.449$ .