Question

Write each expression as a single logarithm. \(\ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right)\)

Short answer

\ln ( \frac{1}{x-1} )

Step-by-step solution

Use the Product Rule

Apply the product rule for logarithms which states that \ \ \ \ \(\ln a + \ln b = \ln (a \cdot b)\) when adding logarithms with the same base. Combine the first two logarithmic expressions: \ \ \ \ \(\begin{aligned} \ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right) = \ln \left(\frac{x}{x-1} \cdot \frac{x+1}{x}\right) \end{aligned}\)

Simplify the Argument

Simplify the argument inside the logarithm by canceling out common factors: \ \ \ \ \ \ \(\begin{aligned} \ln \left( \frac{x}{x-1} \cdot \frac{x+1}{x} \right) &= \ln \left( \frac{x(x+1)}{x(x-1)} \right) = \ln \left( \frac{x +1}{x -1} \right) \end{aligned}\)

Use the Quotient Rule

Apply the quotient rule for logarithms which states that \ \ \ \ \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\) when subtracting logarithms with the same base. Combine the expression from step 2 with the given \ \(\ln \left( x^2 -1 \right)\) term: \ \ \ \ \ \(\begin{aligned} \ln \left( \frac{x+1}{x-1} \right) - \ln (x^2 -1) &= \ln \left( \frac{\frac{x+1}{x-1}}{x^2 -1} \right) \end{aligned}\)

Simplify the Final Expression

Recognize that \ \ \ \ \(x^2 -1\) can be factored as \ (x-1)(x+1). Substitute this in to simplify the logarithmic fraction: \ \ \ \ \ \(\begin{aligned} \ln \left( \frac{\frac{x+1}{x-1}}{(x+1)(x-1)} \right) &= \ln \left( \frac{x+1}{(x-1)(x+1)} \right) \end{aligned}\). \ Finally, cancel out common factors to get \ \ \(\ln ( \frac{1}{x-1} )\).

Key concepts

Product Rule for Logarithms

The product rule for logarithms is essential when you need to combine the sums of logarithmic expressions. It states that: \(\text{ln}(a) + \text{ln}(b) = \text{ln}(a \cdot b)\). This rule helps you convert a sum of logs into a single logarithm. For example, when you see \(\text{ln}(a) + \text{ln}(b)\), you can combine them into one log: \(\text{ln}(a \cdot b)\). Let’s apply this rule to our problem: \(\text{ln}\left(\frac{x}{x-1}\right) + \text{ln}\left(\frac{x+1}{x}\right) = \text{ln}\left(\frac{x}{x-1} \cdot \frac{x+1}{x} \right)\). When you do this, you simplify the operation and make it easier to handle the logarithmic expressions.

Quotient Rule for Logarithms

The quotient rule for logarithms is used when you need to handle the difference of logarithms. It states that: \(\text{ln}(a) - \text{ln}(b) = \text{ln}\left(\frac{a}{b}\right)\). This rule converts a subtraction of two logs into a single logarithm of a fraction. For instance, if you have \(\text{ln}(a) - \text{ln}(b)\), you can merge them into: \(\text{ln}\left(\frac{a}{b}\right)\). In the given exercise, after applying the product rule, we had: \(\text{ln}\left(\frac{x+1}{x-1}\right) - \text{ln}(x^2-1)\). Applying the quotient rule here gives us: \(\text{ln}\left(\frac{\frac{x+1}{x-1}}{x^2-1}\right)\). This transformation simplifies the overall expression.

Simplifying Logarithmic Expressions

Simplifying logarithmic expressions often involves recognizing and canceling out common factors. Let’s take a look into our simplified form: \(\text{ln}\left(\frac{\frac{x+1}{x-1}}{x^2-1}\right)\). Notice that \(x^2 -1 = (x-1)(x+1)\). Realizing this factorization allows you to simplify the argument further. Substituting it in gives us: \(\text{ln}\left(\frac{x+1}{(x+1)(x-1)(x-1)}\right)\). Now, canceling out common factors (the \(x+1\) terms) simplifies it to: \(\text{ln}\left(\frac{1}{x-1}\right)\). This final form, \(\text{ln}\left(\frac{1}{x-1}\right)\), is much simpler and demonstrates how mastering the product and quotient rules, along with recognition of common factors, can elegantly simplify logarithmic expressions.