Question

Solve the following equations for \(x\) : a. \(\cosh (x+\ln 3)=3\) b. \(2 \tanh ^{-1} \frac{x-2}{x-1}=\ln 2\). c. \(\sinh ^{2} x-7 \cosh x+13=0\).

Short answer

\na. \(x = \ln (\frac{-1 \pm 2 \sqrt{10}}{6}) - \ln 3\)\nb. \(x= \frac{3-\sqrt{3}}{2}\)\nc. \(x= \pm acosh (3), \pm acosh (4)\)

Step-by-step solution

Solve \(\cosh(x + \ln 3)=3\)

First, realize that \(\cosh x = e^x + e^{-x} / 2\). The equation implies that \((e^x * e^{\ln 3} + e^{-x} / e^{\ln 3}) / 2 = 3\). Simplify this to get \(3 e^{2x} - 2e^x -3=0\). This quadratic equation can be solved using the Quadratic formula: \(e^{2x} = \frac{2 \pm \sqrt{4 + 4 * 3 * 3}}{2*3}\) or \(e^{2x} = \frac{-1 \pm 2\sqrt{10}}{3}\). To find \(x\), apply \(\ln\) function to both sides, resulting in \(x = \ln (\frac{-1 \pm 2 \sqrt{10}}{6}) - \ln 3\).Note that only the positive root is considered since \(e^x\) is always positive.

Solve \(2 \tanh^{-1} \frac{x-2}{x-1} = \ln 2\)

Use the fact that \(\tanh^{-1} x = 0.5 \ln \frac{1+x}{1-x}\). Substitute to get \( \ln \frac{(x-1)^2 + (x-2)^2}{(x-1)^2 - (x-2)^2} = \ln 2\). Removing the logarithm yields \( \frac{(x-1)^2 + (x-2)^2}{(x-1)^2 - (x-2)^2} = 2\). Solving for \(x\) gives \(x= \frac{3\pm\sqrt{3}}{2}\), noting that only \(x= \frac{3-\sqrt{3}}{2}\) makes sense, as the inverse hyperbolic tangent has the domain \(-1<x<1\).

Solve \(\sinh^2 x - 7 \cosh x + 13=0\)

Start by using the identity \(\sinh^2 x = \cosh^2 x - 1\), so \(\cosh^2 x - 7 \cosh x + 14=0\). Solving this quadratic gives \(\cosh x = \frac{7\pm\sqrt{49-4*14}}{2} = \frac{7\pm\sqrt{1}}{2}\). This results in two solutions: \(\cosh^2 x = 3\) and \(\cosh^2 x = 4\). These give \(x= \pm acosh (3)\) and \(x= \pm acosh (4)\).

Key concepts

Understanding Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest degree term is squared. They generally take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Solving quadratic equations is crucial because they appear frequently in various mathematical contexts.

To find solutions to a quadratic equation, you can use:

• Factoring: Rewriting the equation as a product of binomials, when possible.
• The Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides solutions for any quadratic equation.
• Completing the Square: Transforming the equation into a perfect square trinomial.

In the case of hyperbolic functions, you may encounter quadratic forms like \( e^{2x} \), where solving involves logarithmic operations. For example, in the exercise, solving \( 3e^{2x} - 2e^x - 3=0 \) as a quadratic in terms of \( e^x \) requires using the quadratic formula, and subsequently applying logarithms, as \( e^x \) is an exponent requiring inverse operations to isolate \( x \).

Quadratic roots can be real or complex, and they play a critical role in the curves that the equations describe. Understanding their solutions helps further explore complex relations in functions and equations, especially in advanced topics like hyperbolic functions.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of the hyperbolic functions such as \( \sinh \), \( \cosh \), and \( \tanh \). These functions are often used in calculus and engineering to describe hyperbolic angles and shapes.

Important inverse hyperbolic functions include:

• \( \text{arsinh} \): the inverse of \( \sinh \), expressing as \( \text{arsinh}(x) = \ln(x + \sqrt{x^2 + 1}) \).
• \( \text{arcosh} \): the inverse of \( \cosh \), given by \( \text{arcosh}(x) = \ln(x + \sqrt{x^2 - 1}) \) for \( x \ge 1 \).
• \( \text{artanh} \): the inverse of \( \tanh \), defined as \( \text{artanh}(x) = 0.5 \ln \frac{1+x}{1-x} \), with the domain \( -1 < x < 1 \).

Inverse hyperbolic functions are crucial when addressing problems involving hyperbolic angles or solving certain equations with hyperbolic expressions.

In the exercise, solving \( 2 \tanh^{-1} \frac{x-2}{x-1} = \ln 2 \) involves understanding \( \tanh^{-1} \). This particular expression requires using known logarithmic identities and ensuring that solutions comply with the domain restrictions inherent to \( \tan^{-1} \).

The ability to switch between hyperbolic functions and their inverses opens up solutions to problems involving curves, waveforms, and hyperbolic modeling.

Logarithmic Equations and Their Solutions

Logarithmic equations are equations that involve the logarithm of a variable or an expression. These equations are important because they allow for the transformation and simplification of exponential relationships.

Logarithmic properties are key to solving such equations:

• The product rule: \( \log_b(mn) = \log_b(m) + \log_b(n) \)
• The quotient rule: \( \log_b \left( \frac{m}{n} \right) = \log_b(m) - \log_b(n) \)
• The power rule: \( \log_b(m^k) = k \cdot \log_b(m) \)
• Change of base rule: \( \log_b(m) = \frac{\log_k(m)}{\log_k(b)} \)

Solving logarithmic equations often involves using these rules to simplify and isolate the variable. Additionally, logarithms help reverse the exponential operations, allowing us to express exponents as linear terms.

For example, in the exercise, using the rule \( \ln e^x = x \), we solve equations after transforming expressions like \( 3e^{2x} - 2e^x - 3 = 0 \), by reconfiguring the terms to utilize logarithmic inverses to find \( x \).

Mastering logarithmic equations is essential for mathematical problem-solving, as these concepts connect across algebra, calculus, and beyond. The ability to interrelate them with exponential equations broadens our understanding of mathematical relationships and real-world phenomena.