Problem 5 - Calculus Early Transcendentals

Question

Express \(\sinh ^{-1} x\) in terms of logarithms.

Step-by-Step Solution

Answer

Question: Express the inverse hyperbolic sine function, \(\sinh^{-1} x\), in terms of logarithms. Answer: \(\sinh^{-1} x = \ln[\sqrt{x^2 + 1} + x]\)

Step 1: Definition of Hyperbolic Sine

Recall that the hyperbolic sine function is defined as: $$\sinh(x) = \frac{e^x - e^{-x}}{2}$$

Step 2: Inverse Function

To find the inverse of the function, we will solve for x in terms of y: $$y = \frac{e^x - e^{-x}}{2}$$ We want to rewrite this equation to express \(x\) in terms of \(y\), so we get: $$2y = e^x - e^{-x}$$

Step 3: Simplify and Solve for x

Multiply both sides by \(e^x\) so that we only deal with positive exponents: $$2 ye^x = e^{2x} - 1$$ Now, let \(z = e^x\). We can rewrite the equation as: $$2yz = z^2 - 1$$ Rearrange to get a quadratic equation in terms of z: $$z^2 - 2yz - 1 = 0$$ We can solve this quadratic equation using the quadratic formula, where \(a = 1\), \(b = -2y\), \(c = -1\): $$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Replacing \(a, b, c\) values and simplifying, we get: $$z = \frac{2y \pm \sqrt{4y^2 + 4}}{2}$$ Thus, $$z = y \pm \sqrt{y^2 + 1}$$ Since \(z = e^x\), we can now take the natural logarithm (base e) of both sides to obtain x: $$x = \ln(z) = \ln[\pm \sqrt{y^2 + 1} + y]$$

Step 4: Express Inverse Hyperbolic Sine in terms of Logarithms

Now we can express the inverse hyperbolic sine function \(\sinh^{-1} x\) in terms of logarithms: $$\sinh^{-1} x = \ln[\pm\sqrt{x^2 + 1} + x]$$ However, we need to determine the correct sign. Since \(\sinh(x)\) is an increasing function, its inverse must also be an increasing function. As a result, we must choose the positive square root since only that will result in an increasing function: $$\sinh^{-1} x = \ln[\sqrt{x^2 + 1} + x]$$