Question

Write the following in terms of logarithms: a. \(\cosh ^{-1} \frac{4}{3}\). b. \(\tanh ^{-1} \frac{1}{2}\). c. \(\sinh ^{-1} 2\).

Short answer

a. \(\cosh ^{-1} \frac{4}{3} = \ln(2 + \sqrt{3})\)\nb. \(\tanh ^{-1} \frac{1}{2} = \ln(\sqrt{3})\)\nc. \(\sinh ^{-1} 2 = \ln(2 + \sqrt{5})\)

Step-by-step solution

Converting inverse hyperbolic function into algebraic expressions

a. Let \(y = \cosh ^{-1} \frac{4}{3}\). Hence, \(\cosh y = \frac{4}{3}\).\nb. Let \(y = \tanh ^{-1} \frac{1}{2}\). Hence, \(\tanh y = \frac{1}{2}\).\nc. Let \(y = \sinh ^{-1} 2\). Hence, \(\sinh y = 2\).

Rewrite them in terms of exponentials

a. For \( \cosh y = \frac{4}{3}\), we get \( \cosh y = \frac{e^y + e^{-y}}{2} = \frac{4}{3}\).\nb. For \( \tanh y = \frac{1}{2}\), we get \( \tanh y = \frac{e^y - e^{-y}}{e^y + e^{-y}} = \frac{1}{2}\).\nc. For \( \sinh y =2 \), we get \(\sinh y = \frac{e^y - e^{-y}}{2} = 2\).

Solve these expressions

a. The equation \( \frac{e^y + e^{-y}}{2} = \frac{4}{3}\) simplifies to \(e^y + e^{-y} = \frac{8}{3}\).\nb. The equation \( \frac{e^y - e^{-y}}{e^y + e^{-y}} = \frac{1}{2}\) simplifies to \(e^y - e^{-y}=e^y + e^{-y}\).\nc. The equation \( \frac{e^y - e^{-y}}{2} =2\) simplifies to \(e^y - e^{-y} = 4\). After these simplifications, the equations can be rewritten in terms of logarithms.

Rewrite them in terms of logarithms

a. After simplification, for \(\cosh ^{-1} \frac{4}{3}\) we get \(y = \ln(2 + \sqrt{3})\).\nb. After simplification, for \(\tanh ^{-1} \frac{1}{2}\) we get \(y = \ln(\sqrt{3})\).\nc. After simplification, for \(\sinh ^{-1} 2\) we get \(y = \ln(2 + \sqrt{5})\).

Key concepts

Cosh Inverse

The hyperbolic cosine inverse, denoted as \( \cosh^{-1} x \), is one of the inverse hyperbolic functions. To understand it better, it helps to first know that the hyperbolic cosine function is defined as \( \cosh x = \frac{e^{x} + e^{-x}}{2} \). Therefore, \( \cosh^{-1} x \) gives us the value of y such that \( x = \cosh y \).

In algebraic terms, if \( x = \cosh y = \frac{e^y + e^{-y}}{2} \), then we can solve for y by setting up an exponential equation and rearranging it to get \( y = \ln(x + \sqrt{x^2 - 1}) \), provided that \( x \) is greater than or equal to 1. This is derived from the quadratic formula application to the variable \( e^y \) after moving all other terms to the opposite side of the equation.

Tanh Inverse

Inverse hyperbolic tangent, or \( \tanh^{-1} x \), is another inverse hyperbolic function which finds the value y corresponding to the hyperbolic tangent of x. The formula for the hyperbolic tangent is \( \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \).

To express \( \tanh^{-1} x \) in terms of logarithms, we need to solve an algebraic equation as \( x = \tanh y \), which simplifies to \( y = \frac{1}{2} \ln(\frac{1 + x}{1 - x}) \), where \( -1 < x < 1 \). This identity is particularly useful when integrating certain types of functions or solving equations in calculus.

Sinh Inverse

The inverse hyperbolic sine, \( \sinh^{-1} x \), relates to \( x \) in a way that \( x = \sinh y \) where \( y \) is obtained from the expression \( \sinh y = \frac{e^y - e^{-y}}{2} \).

To convert \( \sinh^{-1} x \) into logarithmic form, we manipulate the equation \( x = \sinh y \) resulting in \( y = \ln(x + \sqrt{x^2 + 1}) \). This conversion is essential in various areas including differential equations and complex analysis.

Exponential Equations

Exponential equations involve variables in the exponent. The most common base for these types of equations in calculus and hyperbolic function problems are the natural base, e. An exponential equation like \( e^y - e^{-y} = z \) can be transformed into a logarithmic equation using properties of logarithms.

Solving these requires manipulating the equation to isolate the term with the exponent first. If \( e^y \) and \( e^{-y} \) are on opposite sides of the equation, a common strategy is to multiply by \( e^y \) to rid the equation of the negative exponent, making it easier to isolate \( y \) and convert it to logarithmic form.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation symbols. When dealing with inverse hyperbolic functions and their conversions into logarithms, algebraic manipulation is key.

Students should be comfortable with standard algebraic techniques such as factoring, expanding, and using the quadratic formula. These skills are crucial when working to isolate the variable and translate the expression into its equivalent logarithmic form. For example, to solve for y in exponential equations, one often needs to perform a series of algebraic transformations, setting the stage for the final step of employing the logarithmic functions to find the solution.