Problem 10 - Thomas Calculus

Question

In Exercises \(5-10,\) solve for \(y\) in terms of \(t\) or \(x,\) as appropriate. $$ \ln \left(y^{2}-1\right)-\ln (y+1)=\ln (\sin x) $$

Step-by-Step Solution

Answer

The solutions for \(y\) are \(y = \sin x + 1\) and \(y = -1\).

Step 1: Use the Properties of Logarithms

We start by using the property of logarithms that allows us to combine or simplify them: \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\). Thus, \(\ln(y^2 - 1) - \ln(y + 1) = \ln\left(\frac{y^2 - 1}{y + 1}\right)\). The equation now becomes:\[\ln\left(\frac{y^2 - 1}{y + 1}\right) = \ln(\sin x)\]

Step 2: Exponentiate to Remove the Logarithms

To eliminate the natural logarithm, we exponentiate both sides of the equation. This gives: \[\frac{y^2 - 1}{y + 1} = \sin x\]

Step 3: Solve for y by Clearing the Fraction

Multiply both sides by \(y + 1\) to clear the fraction: \[y^2 - 1 = (y + 1)\sin x\]

Step 4: Expand and Rearrange the Equation

Expand the right-hand side: \(y^2 - 1 = y\sin x + \sin x\). Rearrange terms to isolate \(y\) terms: \[y^2 - y\sin x = \sin x + 1\]

Step 5: Recognize a Quadratic Equation in y

The equation \(y^2 - y\sin x - (\sin x + 1) = 0\) is a quadratic equation in terms of \(y\).

Step 6: Apply the Quadratic Formula

The quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -\sin x\), and \(c = -(\sin x + 1)\). Substitute these into the formula: \[y = \frac{\sin x \pm \sqrt{(\sin x)^2 + 4(\sin x + 1)}}{2}\]

Step 7: Simplify the Expression

Simplify the expression under the square root: \(b^2 - 4ac = (\sin x)^2 + 4\sin x + 4\), and thus the equation becomes: \[y = \frac{\sin x \pm \sqrt{(\sin x + 2)^2}}{2}\]. Simplifying further gives:\[y = \frac{\sin x \pm (\sin x + 2)}{2}\]

Step 8: Solve for y

This splits into two possible expressions for \(y\):\[y = \frac{\sin x + \sin x + 2}{2} = \sin x + 1\]\[y = \frac{\sin x - (\sin x + 2)}{2} = -1\]

Key Concepts

Properties of Logarithms

When solving logarithmic equations, the properties of logarithms are essential tools. One key property is the difference rule: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This property allows us to combine two logarithms into a single logarithm, making calculations simpler and more straightforward. For example, in our equation, we had \( \ln(y^2 - 1) - \ln(y + 1) \), which we turned into \( \ln\left(\frac{y^2 - 1}{y + 1}\right) \) using this property.

• This simplification helps to unify the logarithmic terms on one side of the equation.
• Applying properties like this is fundamental for solving complex logarithmic expressions more efficiently.

Understanding these properties thoroughly can simplify many problems you encounter.

Quadratic Formula

The quadratic formula is a powerful tool in algebra for solving equations that follow this structure: \( ax^2 + bx + c = 0 \). It's expressed as: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In the context of our problem, we recognized a quadratic form: \( y^2 - y\sin x - (\sin x + 1) = 0 \). Here, \( a = 1 \), \( b = -\sin x \), and \( c = -(\sin x + 1) \).

• We use the formula to find values of \( y \) that satisfy the equation.
• The \( \pm \) sign indicates there are typically two solutions.

Calculating these helps us to uncover possible values for unknowns in equations of this nature.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm with the base \( e \), where \( e \approx 2.718 \). It's widely used in mathematics, especially in calculus and complex calculations. In the context of our original equation, natural logarithms helped simplify expressions by using their properties.
Understanding how \( \ln(x) \) behaves:

• It grows logarithmically, meaning it increases slowly for larger values of \( x \).
• It converts multiplicative relationships into additive ones, aiding in solving equations.

Familiarity with \( \ln(x) \) can greatly enhance problem-solving skills in interference with exponential growth and decay models.

Exponentiating Equations

Exponentiating is an essential step when solving equations involving logarithms. When you have an equation with natural logarithms like \( \ln(a) = \ln(b) \), we can remove the logarithms by exponentiating: if \( \ln(a) = \ln(b) \), then \( a = b \). This principle applies because exponentiation is the inverse operation of taking a logarithm.

• Exponentiating allows us to simplify and directly solve the original problem.
• For our equation, after exponentiating, we had \( \frac{y^2 - 1}{y + 1} = \sin x \), a clearer relationship connecting \( y \) and \( x \).

Such techniques are invaluable for transitioning from logarithmic equations to more elementary forms that are often easier to work with analytically.