Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x+\ln (x-2)=1\)

Short answer

The solution for the given logarithmic equation is approximately \(x \approx 2.718\). Note: Only positive solutions are valid for logarithmic equations.

Step-by-step solution

Combine the log terms

Use the property of logarithms \(\ln a + \ln b = \ln (ab)\) to combine the log terms. This results to: \(\ln (x*(x-2)) = 1\)

Exponentiate both sides

To get rid of the natural logarithm, the inverse function, the exponentiation to the base e, can be applied. After applying the inverse function, the equation becomes: \(x*(x-2) = e^1\)

Simplify the equation

Simplify the equation by distributing \(x\) on the left side and evaluating \(e^1\) on the right side. This results in: \(x^2 - 2x = e\)

Rearrange the equation into quadratic form

To configure the cubic equation into a quadratic form, it should be set equal to zero. This results into: \(x^2 - 2x - e = 0\)

Solve the quadratic equation

Now, solve the quadratic equation using the quadratic formula \(x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). After substituting \(a = 1\), \(b = -2\), and \(c = -e\) into the quadratic formula, it results into two solutions for 'x': \(x_1 = 1 + \sqrt{1 + e}\) and \(x_2 = 1 - \sqrt{1 + e}\)

Check for positive solutions

In the context of the natural logarithm, only positive solutions are valid. Checking the solutions, \(x_1 = 1 + \sqrt{1 + e}\) is positive while \(x_2 = 1 - \sqrt{1 + e}\) is negative. Therefore, discard \(x_2\).

Approximate the solution

Use a calculator to approximate the value of \(x\) to three decimal places: \(x \approx 2.718\)

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \(ax^2 + bx + c = 0\). This type of equation is called a quadratic equation, where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. The quadratic formula is given as:\[x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]When using the quadratic formula, it's important to find the values of \(a\), \(b\), and \(c\) from the original equation first:

• \(a\) is the coefficient of \(x^2\).
• \(b\) is the coefficient of \(x\).
• \(c\) is the constant term or intercept.

The discriminant, \(b^2 - 4ac\), determines how many solutions there are:

• If it is positive, there are two distinct solutions.
• If it is zero, there is one real solution.
• If it is negative, the solutions are complex numbers.

In the original problem, the quadratic formula was used after rewriting the equation to the form \(x^2 - 2x - e = 0\), leading to one valid solution for \(x\). This solution is crucial in solving logarithmic equations, especially when transitioning from a logarithmic form to quadratic form.

Natural Logarithm

The natural logarithm is a special logarithm which uses the mathematical constant \(e\) (approximately 2.718) as its base. It is denoted as \(\ln\). The natural logarithm has unique properties that make it incredibly useful in calculus, physics, and other applied sciences. One key property is the power of simplification during multiplication and division of exponential terms:- \(\ln (ab) = \ln a + \ln b\)- \(\ln \left(\frac{a}{b}\right) = \ln a - \ln b\)In solving logarithmic equations, properties like \(\ln a + \ln b = \ln (ab)\) allow combining terms effectively. For instance, in the exercise \(\ln x + \ln (x-2) = 1\), these logarithmic properties simplify the equation into a more manageable form: \(\ln(x(x-2)) = 1\).Moreover, solving such equations often involves inverting the logarithmic function, using exponentiation, to remove \(\ln\) from the equation, leading to an exponential equation easily solved by other algebraic methods.

Exponentiation

Exponentiation is a fundamental operation in mathematics that deals with raising numbers to powers. Essentially, when a number \(a\) is raised to a power \(b\), it is expressed as \(a^b\). This concept is crucial when working with logarithms. Specifically, in natural logarithms, the inverse operation involves exponentiation with base \(e\).During the solving of logarithmic equations, once logarithmic terms are combined or simplified, exponentiation is used to isolate the variable. Take for example the equation \(\ln(x(x-2)) = 1\). By exponentiating both sides with base \(e\), you transform and eliminate the natural log:\[e^{\ln(x(x-2))} = e^1\]This simplifies to:\[x(x-2) = e\]By exponentiating, you seamlessly transition from a logarithmic expression to a polynomial equation, which can then be resolved through methods such as factoring or applying the quadratic formula. Understanding exponentiation helps in bridging different types of equations, leading to comprehensive solutions in algebra.