Question

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

Short answer

Question: Show that \(\ln \left(\frac{x}{y}\right) = \ln x - \ln y\) using the integral definition of the natural logarithm. Answer: \(\ln \left(\frac{x}{y}\right) = \ln x - \ln y\)

Step-by-step solution

Recall the integral definition of the natural logarithm

The natural logarithm of a positive number \(x\) can be defined as the integral: \[ \ln x = \int_1^x \frac{1}{t} dt \] Similarly, for the number \(y\), we have: \[ \ln y = \int_1^y \frac{1}{t} dt \]

Find the integral for the natural logarithm of \(x/y\)

Now we need to find the integral expression for the natural logarithm of the quantity \(\frac{x}{y}\). By using the integral definition of the natural logarithm, we have: \[ \ln \left(\frac{x}{y}\right) = \int_1^{\frac{x}{y}} \frac{1}{t} dt \]

Perform substitution to match the limits of the integral for ln(x)

To compare the ln expressions, we need to match the limits of the integral for ln(x). To do this, we can use the substitution \(u = \frac{t}{y}\). Then \(t=uy\), and \(\frac{1}{t} = \frac{1}{uy}\). The differentials are related by \(dt = y du\). We also need to change the limits of integration. When \(t = 1\), \(u = \frac{1}{y}\), and when \(t =\frac{x}{y}\), \(u = x\). Now the integral becomes: \[ \ln \left(\frac{x}{y}\right) = \int_\frac{1}{y}^x \frac{1}{uy} y du = \int_\frac{1}{y}^x \frac{1}{u} du \]

Split the integral expression and apply the limits of integration

Now we will split the integral of \(\ln \left(\frac{x}{y}\right)\) into two separate integrals: \[ \ln \left(\frac{x}{y}\right) = \int_\frac{1}{y}^x \frac{1}{u} du = \int_\frac{1}{y}^1 \frac{1}{u} du + \int_1^x \frac{1}{u} du \] Notice that the second integral is simply the expression for \(\ln x\). To find the expression for the first integral, we will again use a substitution. Let \(v = \frac{1}{u}\). Then, \(u = \frac{1}{v}\) and \(\frac{1}{u} = v\). The differentials are related by \(du = -\frac{1}{v^2} dv\). We also need to change the limits of integration. When \(u =\frac{1}{y}\), \(v = y\). When \(u =1\), \(v =1\). Now the integral becomes: \[ \int_\frac{1}{y}^1 \frac{1}{u} du = \int_y^1 -\frac{1}{v^2} dv \]

Show that the sum of the integrals equals ln(x) - ln(y)

Adding the two integral expressions from Step 4, we have: \[ \ln \left(\frac{x}{y}\right) = \int_y^1 -\frac{1}{v^2} dv + \int_1^x \frac{1}{u} du \] Now, we can rewrite the first integral to have limits of integration going from 1 to y. To do this, swap the limits of integration and remember to change the sign of the function being integrated: \[ \ln \left(\frac{x}{y}\right) = \int_{1}^y \frac{1}{v^2} dv + \int_1^x \frac{1}{u} du \] This sum of integrals is equal to: \[ \ln \left(\frac{x}{y}\right) = \int_1^y \frac{1}{v^2} dv + \int_1^x \frac{1}{u} du = \ln x - \ln y \] Therefore, we have shown using the integral definition of the natural logarithm that \(\ln \left(\frac{x}{y}\right) = \ln x - \ln y\).

Key concepts

Integral Definition

The integral definition of the natural logarithm is a powerful tool to understand the logarithmic function and its properties. At its core, the natural logarithm of a number, say x, can be expressed as an integral from 1 to x of the reciprocal function: \[ \ln x = \int_1^x \frac{1}{t} dt\]This integral definition provides an essential insight into the nature of the logarithm as a continuous function. Through this integral, you can see how the accumulation of areas under the curve of the function \( \frac{1}{t} \) provides the value of \( \ln x \). This representation encourages a deeper understanding of how logarithms behave over different intervals.
When using the integral definition, it's crucial to remember these basic principles:

• The integration starts from 1 because \( \ln 1 = 0 \), marking the base point for natural logarithms.
• The denominator \( \frac{1}{t} \) reminds us that the natural logarithm is concerned with the growth of the function as the variable t moves.

Logarithmic Properties

The properties of logarithms help simplify complex mathematical expressions and solve logarithmic equations efficiently. One such property is the quotient rule, which states:\[\ln \left(\frac{x}{y}\right) = \ln x - \ln y\]This property is derived directly using the integral definition of the natural logarithm. By considering \[\ln \left(\frac{x}{y}\right) = \int_1^{\frac{x}{y}} \frac{1}{t} dt \]we can decompose the integral through substitution, allowing us to separate it into two integrals that correspond to \( \ln x \) and \(-\ln y \) respectively.
The quotient rule for logarithms is useful because:

• It simplifies division inside the logarithm into a subtraction, easing calculations.
• It's derived and validated through integral transformation and substitution techniques.
• It maintains the properties of continuity and differentiability, which are intrinsic to natural logarithms.

Limits of Integration

When evaluating the integral expression for the natural logarithm, limits of integration are crucial. They define the range over which the function \( \frac{1}{t} \) is integrated, ultimately influencing the resulting value of the logarithm. Take, for example, the expression\[\ln \left( \frac{x}{y} \right) = \int_1^{\frac{x}{y}} \frac{1}{t} dt \]To prove reliably that this equals \( \ln x - \ln y \), we adjust limits through substitution. The limits originally go from 1 to \( \frac{x}{y} \), but by carefully choosing a substitution like \( u = \frac{t}{y} \), we change these integration limits from \( \frac{1}{y} \) to x. This substitution simplifies the expression, helping us split and evaluate it to show the original identity holds true.
Key points to remember about limits of integration:

• The choice of limits determines the area under the curve, affecting the natural logarithm's value.
• Through substitution, limits can be adjusted to facilitate simpler calculations and transformations.
• Limits must maintain the integrity of the original function's range and behavior.