Question

Prove that \(\ln (x / y)=\ln x-\ln y, \quad x>0, y>0\).

Short answer

The identity \(\ln (x / y)=\ln x-\ln y\) has been proven by using the properties of logarithms and converting between logarithmic and exponential forms.

Step-by-step solution

Recall Properties of Logarithms

Let's recall the property of logarithms that states that the logarithm of a quotient is equal to the difference of the logarithms: \(log_b \frac{a}{c} = log_b a - log_b c\). The same property applies for natural logarithms and will be used in the proof.

Establish Identity to Prove

Let's now establish the identity we need to prove, \( \ln (x / y)=\ln x-\ln y\), and set as \(z = \ln x - \ln y\). By the properties of logarithms we know that this is equal to \(z = \ln (x / y)\)

Convert the Natural Logarithm to Exponential Form

Next, let's convert the natural logarithm \(z = \ln (x / y)\) to exponential form. We get \(e^z = e^{\ln (x / y)}\). Using the property that \(e^{\ln a} = a\) for any positive number a, this simplifies to \(e^z = x / y\). Also, from \(z = \ln x - \ln y\),using properties of exponents, we can rewrite this as \(e^z = e^{\ln x} / e^{\ln y}\). Again, using the property that \(e^{\ln a} = a\) for any positive number a, this simplifies to \(e^z = x / y\).

Equate and Conclude

From steps 3, we are able to conclude that both expressions \(e^z = x / y\) are equal. Thus, we've indeed verified that \(\ln (x / y) = \ln x - \ln y\).