Chapter 3.4, Problem 62E

To determine

Find the solution of the given logarithmic equation algebraically.

Answer to Problem 62E

The solution of the equation is $x=180.384$ .

Explanation of Solution

Given:

The given equation is $\log 8x-\log \left(1+\sqrt{x}\right)=2$ .

Calculation:

  $\log 8x-\log \left(1+\sqrt{x}\right)=2$

Use quotient property.

  ${\log }_{p}q-{\log }_{p}r={\log }_p\left(\frac{q}{r}\right)$

  $\log \left(\frac{8x}{1+\sqrt{x}}\right)=2$

If $\ln q=r$ , then $q=e^r$ .

  $\begin{array}{l}\Rightarrow \frac{8x}{1+\sqrt{x}}={10}^2\\ \Rightarrow \frac{8x}{1+\sqrt{x}}=100\end{array}$

Multiply both sides by $1+\sqrt{x}$ .

  $\begin{array}{l}\Rightarrow \frac{8x}{\left(1+\sqrt{x}\right)}\left(1+\sqrt{x}\right)=100\left(1+\sqrt{x}\right)\\ \Rightarrow 8x=100+100\sqrt{x}\end{array}$

Subtract $100$ from both sides.

  $\begin{array}{l}\Rightarrow 8x-100=100+100\sqrt{x}-100\\ \Rightarrow 8x-100=100\sqrt{x}\end{array}$

Squaring both sides.

  $\begin{array}{l}\Rightarrow 64x^2+10000-1600x=10000x\\ \Rightarrow 64x^2-11600x+10000=0\\ \Rightarrow 4x^2-725x+625=0\end{array}$

Use quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

  $a=4,b=-725,c=625$

  $\begin{array}{l}x=\frac{-\left(-725\right)\pm \sqrt{{\left(-725\right)}^2-4\cdot 4\cdot \left(625\right)}}{2\cdot 4}\\ \Rightarrow x=\frac{725\pm \sqrt{525625-10000}}{8}\\ \Rightarrow x=\frac{725\pm \sqrt{515625}}{8}\\ \Rightarrow x=\frac{725\pm 718.0703}{8}\\ \Rightarrow x=\frac{1443.0703}{8}\text{and}\Rightarrow x=\frac{-6.9296}{8}\\ \Rightarrow x=180.384\text{and}\Rightarrow x=-0.8662\end{array}$

Verify the result.

  $\begin{array}{l}\log 8\left(180.384\right)-\log \left(1+\sqrt{180.384}\right)\approx 2\\ \log 8\left(-0.8662\right)-\log \left(1+\sqrt{-0.8662}\right)\ne 2\end{array}$

Hence thesolution of the equation is $x=180.384$ .