Chapter 3.4, Problem 56E

To determine

To find : solution to the logarithmic equation algebraically.

Answer to Problem 56E

The solution to the equation $\ln x+\ln \left(x+1\right)=1$ is $\underset{\_}{\boxed{x\approx 1.223}}$

Explanation of Solution

Given information: $\ln x+\ln \left(x+1\right)=1$

Concept Involved:

The word “solve” means the process of finding the value of t that makes the equation true. In order to solve we need to undo whatever is done to t .

Formula Used

1. $\ln m+\ln n=\ln \left(m\cdot n\right)$
2. If logarithmic equation is $\ln A=m$ , then the equivalent exponential equation is $A=e^m$

Calculation:

Use the logarithmic rule $\ln m+\ln n=\ln \left(m\cdot n\right)$ in left side to rewrite the equation

  $\ln \left(x\cdot \left(x+1\right)\right)=1$

Rewrite the logarithmic equation to exponential equation

  $x\left(x+1\right)=e^1$

Distribute x in left side of the equation

  $x^2+x=e$

Subtract $e$ on both sides of the equation

  $x^2+x-e=e-e$

Simplify on right side of the equation

  $x^2+x-e=0$

When the quadratic equation of the form $a{x}^2+bx+c=0$ , we can use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

Find the two values of x by simplifying further

$x=\frac{-1+\sqrt{1+4e}}{2}$$x\approx 1.22287$

$x=\frac{-1-\sqrt{1+4e}}{2}$$x\approx -2.22287$

Check for extraneous solution by substituting the result in the original equation.

  $x\approx -2.22287$ makes the original equation FALSE&

  $x\approx 1.22287$ makes the original equation TRUE.

Conclusion:

The solution to the equation $\ln x+\ln \left(x+1\right)=1$ is $\underset{\_}{\boxed{x\approx 1.223}}$