Problem 42 - Calculus Early Transcendentals

Question

Use the integral definition of the natural logarithm to prove that \(\ln (x / y)=\ln x-\ln y\).

Step-by-Step Solution

Answer

Question: Prove that the logarithmic identity \(\ln(x/y) = \ln x - \ln y\) holds based on the integral definition of the natural logarithm. Solution: We have shown that by using the integral definition of the natural logarithm and performing algebraic manipulations, we can arrive at the desired result: \(\ln(x/y) = \ln x - \ln y\). The steps followed include recalling the integral definition, rewriting the expression, transforming the limits of integration, splitting the integral, using the integral definition, and lastly obtaining the final expression for the logarithmic identity.

Step 1: Recall the integral definition of the natural logarithm

The natural logarithm function, denoted as ln(x), can be defined as the integral of the reciprocal function, from 1 to x, with respect to t: \[ \ln(x) = \int_{1}^{x} \frac{1}{t} \ dt \]

Step 2: Write the expression for \(\ln(x/y)\) using the integral definition

Using the integral definition of the natural logarithm, we can write the expression for \(\ln(x/y)\) as: \[ \ln\left(\frac{x}{y}\right) = \int_{1}^{\frac{x}{y}} \frac{1}{t} \ dt \]

Step 3: Transform the limits of integration and the integrand

We can make a substitution \(u = \frac{t}{y}\), then \(t = uy\) and \(dt = y \ du\). The limits of integration will transform as well, with the new limits being: \[ \begin{cases} u = \frac{1}{y} & \text{when} \ t=1 \\ u = \frac{x}{y} & \text{when} \ t=\frac{x}{y} \end{cases} \] Now we have: \[ \ln\left(\frac{x}{y}\right) = \int_{\frac{1}{y}}^{\frac{x}{y}} \frac{y}{uy} \ dy \]

Step 4: Split the integral

Now, we can split the integral into two parts by setting an intermediate limit at \(\frac{1}{y}+\frac{x}{y}\): \[ \ln\left(\frac{x}{y}\right) = \int_{\frac{1}{y}}^1 \frac{y}{uy} \ dy + \int_{1}^{\frac{x}{y}} \frac{y}{uy} \ dy \]

Step 5: Use the integral definition of the natural logarithm

The integrals obtained in the previous step can be rewritten using the integral definition of the natural logarithm, as explained in Step 1: \[ \ln\left(\frac{x}{y}\right) = - \int_{1}^{\frac{1}{y}} \frac{1}{u} \ du + \int_{1}^{\frac{x}{y}} \frac{1}{u} \ du \]

Step 6: Obtain the final expression

In this final step, we apply the fundamental theorem of calculus to evaluate the two integrals and obtain the desired result: \[ \ln\left(\frac{x}{y}\right) = - \ln\left(\frac{1}{y}\right) + \ln\left(\frac{x}{y}\right) \] As we know the properties of logarithms, \( - \ln\left(\frac{1}{y}\right) = \ln y\). Therefore, we have: \[ \ln\left(\frac{x}{y}\right) = \ln x - \ln y \] This concludes the proof of the logarithmic identity \(\ln(x/y) = \ln x - \ln y\).