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Chapter 3: Problem 100

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln x+\ln (x+1)=1$$

Short Answer

Expert verified
The solution to the logarithmic equation \( \ln x + \ln (x+1) = 1 \) is \( x = \frac{-1 + \sqrt{1 + 4e}}{2} \)

Step by step solution

01

Combine the logarithms

With the use of logarithmic properties, the left-hand side of the equation, which contains two logs added together, can be combined into a single log. According to the properties of logs, \(\ln a + \ln b = \ln (a*b) \). Hence, \(\ln x + \ln (x+1) = \ln (x*(x+1)) \). So, the equation becomes: \(\ln (x*(x+1)) = 1 \).
02

Convert to exponential form

Getting rid of the ln by using the property that shows the relationship between logarithmic form and exponential form, the equation can be rewritten. The natural log has a base \( e \), so the equation, when expressed in exponential form, will be: \( e^1 = x*(x+1) \). Thus, the equation becomes: \( e = x*(x+1) \).
03

Expand and rearrange into quadratic form

By expanding \( x * (x + 1) \), we get \( x^2 + x \). This makes our equation look like a quadratic equation: \( x^2 + x - e = 0 \).
04

Solve the quadratic equation

This equation needs to be solved for \( x \). The quadratic formula is \( x = [-b \pm \sqrt{b^2 - 4ac} ] / (2a) \). Here \( a=1 \), \( b=1 \), \( c=-e \)
05

Calculate the value of x

Substituting \( a=1 \), \( b=1 \), \( c=-e \) into quadratic formula we get two potential solutions for \( x \) which are \( x_1 = \frac{-1 + \sqrt{1 + 4e}}{2} \) and \( x_2 = \frac{-1 - \sqrt{1 + 4e}}{2} \). However, based on the original equation we know that \( x > 0 \) and \( x + 1 > 0 \), and thus, \( x_2 < 0 \) can be rejected as a solution.

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Most popular questions from this chapter

The demand equation for a hand-held electronic organizer is \(p=5000\left(1-\frac{4}{4+e^{-0.002 x}}\right)\) Find the demand \(x\) for a price of (a) \(p=\$ 600\) and (b) \(p=\$ 400\).

Use the acidity model given by \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right],\) where acidity \((\mathrm{pH})\) is a measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Compute \(\left[\mathrm{H}^{+}\right]\) for a solution in which \(\mathrm{pH}=3.2\).

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Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$-x e^{-x}+e^{-x}=0$$

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