Problem 56 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

After solving, we find that \(x\) approximately equals 0.567 when rounded off to three decimal places.

Step 1: Combine the logarithms

We start off by applying the rule of logarithms \(ln a + ln b = ln(a*b)\), which allows us to combine the logs. This gives us:\(\ln(x*(x+1)) = 1\)

Step 2: Convert from log form to exponential form

Next we convert from logarithmic form to exponential form. The relationship between logarithmic and exponential forms is as follows: if we have \(\ln a = b\), we can write this as \(e^b = a\). Converting gives us the equation \(e^1 = x*(x+1)\). Hence, \(e = x*(x+1)\)

Step 3: Convert to a Quadratic Equation

In trying to solve for 'x', we realize that this equation is a quadratic equation. So we set it equal to zero: \(x^2 + x - e = 0\)

Step 4: Solve the Quadratic Equation

Now we solve for 'x'. For a quadratic equation of the form \(ax^2 + bx + c = 0\), the solutions are given by \(x = [-b \pm sqrt(b^2 - 4ac)] / (2a)\). Substituting the values \(a=1\), \(b=1\) and \(c=-e\), we solve for 'x'. Due to the nature of logs, we disregard the negative solution and only consider the positive one. Remember to approximate to three decimal places.

Key Concepts

Quadratic Equations

Quadratic equations represent a fundamental concept in algebra. These equations are usually of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). Let's break it down:

• Standard Form: The "\(ax^2\)" term makes it quadratic, meaning the highest power of the variable \(x\) is 2.
• Roots: Solving a quadratic equation involves finding the roots, which are the values of \(x\) that make the equation true.
• Quadratic Formula: Typically, we solve these equations using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula allows us to calculate the roots directly.

Remember, in our original exercise, after transforming the exponential form into a quadratic equation, we applied this formula to solve for \(x\). The result gave us the possible values of \(x\), but only the positive root was considered because logarithms can't have negative arguments or zero. So, it's important to check solutions against the domain restrictions of the original problem.

Logarithmic Functions

Logarithmic functions are essential for solving equations involving exponential relationships. A logarithmic function is the inverse of an exponential function. Here's what you need to know:

• Basic Form: Expressed as \(y = \log_b(x)\), where \(b\) is the base. For natural logarithms, we use \(y = \ln x\), meaning the base \(b\) is Euler's number \(e\), approximately 2.718.
• Properties: Logarithms have useful properties that help simplify equations, such as \(\ln(ab) = \ln a + \ln b\). These properties were used in our original exercise to combine logarithms for simplification.
• Domain: The domain of a logarithmic function is all positive real numbers. This aspect restricts the solutions we accept for equations involving logarithms.

In the original problem, we combined two logarithmic terms into one using the property \(\ln a + \ln b = \ln(ab)\). Then, we used the fact that \(\ln a = b\) translates to \(a = e^b\) to re-write the logarithmic equation as an exponential one.

Exponential Functions

Exponential functions are characterized by their rapid growth or decay. They are generally written as \(f(x) = a \, b^x\) and have the following aspects:

• Function Form: Typically shown as \(y = a^x\), where \(a > 0\), \(y\) increases exponentially as \(x\) increases if \(a > 1; \) decreases if \(0 < a < 1\).
• Relation to Logarithms: Exponential functions are the inverse of logarithmic functions. This relationship means you can switch between the two forms easily. For example, \(e^y = x\) translates to \(y = \ln x\).
• Applications: Commonly used in fields such as science, finance, and statistics to model various phenomena like population growth, radioactive decay, and compound interest.

In our original solution, we converted the logarithmic equation into an exponential one. This step involved recognizing that if we have \(\ln(xy) = 1\), we can express it as \(e^1 = xy\), thus transitioning from the logarithmic to exponential form. Understanding this transition is crucial for solving many logarithm-based problems efficiently.