Problem 110 - Calculus Early Transcendentals

Question

Inverse hyperbolic tangent Recall that the inverse hyperbolic tangent is defined as \(y=\tanh ^{-1} x \Leftrightarrow x=\tanh y,\) for \(-1

Step-by-Step Solution

Answer

Question: Express the inverse hyperbolic tangent function, \(y = \tanh^{-1} x\), in terms of logarithms. Answer: \(y = \ln{\frac{-(-1 + 2x) \pm \sqrt{(-1 + 2x)^2 - 4(x - 1)(x)}}{2(x - 1)}}\)

Step 1: Introduce the definition of the hyperbolic tangent

Let's start with the definition of the hyperbolic tangent function, which is given by: \(\tanh y = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}\) Our given equation is \(x = \tanh y\). Therefore, we can write: \(x = \frac{e^y - e^{-y}}{e^y + e^{-y}}\) Now, our goal is to solve for \(y\) in terms of \(x\) and logarithms.

Step 2: Cross-multiply and rearrange terms

Cross-multiply \(x = \frac{e^y - e^{-y}}{e^y + e^{-y}}\) and then rearrange the equation to get \(x * (e^y + e^{-y}) = e^y - e^{-y}\). The equation now becomes: \(x(e^y + e^{-y}) = e^y - e^{-y}\)

Step 3: Add product terms to remove the negative exponents

To remove the negative exponents \(e^{-y}\), we will add \(2x * (e^y)(e^{-y})\) to both sides of the equation. This gives us: \(x(e^y + e^{-y}) + 2x * (e^y)(e^{-y}) = e^y - e^{-y} + 2x * (e^y)(e^{-y})\) Simplify the left side of the equation: \(x(e^y + e^{-y} + 2x*(e^y)(e^{-y})) = e^y + (2x - 1) * e^{-y}\)

Step 4: Make a substitution and isolate \(y\)

Let's make a substitution, \(z = e^y\). Therefore, the equation becomes: \(x(z + \frac{1}{z} + 2x * z * \frac{1}{z}) = z + (2x - 1) \frac{1}{z}\) After simplifying, we get the following quadratic equation in \(z\): \(x(z^2 + 1 + 2xz) - z^2 + z(2x - 1) = 0\) Now solve for \(z\): \((x - 1)z^2 + (-1 + 2x)z + x = 0\)

Step 5: Solve the quadratic equation for \(z\)

Use the quadratic formula to solve for \(z = e^y\): \(z = \frac{-(-1 + 2x) \pm \sqrt{(-1 + 2x)^2 - 4(x - 1)(x)}}{2(x - 1)}\)

Step 6: Apply inverse logarithm operation

Now that we have \(z = e^y\), apply the logarithm function to both sides of the equation to isolate \(y\): \(y = \ln{z}\) Substitute the expression for \(z\) we found before: \(y = \ln{\frac{-(-1 + 2x) \pm \sqrt{(-1 + 2x)^2 - 4(x - 1)(x)}}{2(x - 1)}}\) This is the expression for the inverse hyperbolic tangent function, \(y = \tanh^{-1} x\), in terms of logarithms.