Problem 83 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the properties of logarithms to condense the expression.$$\frac{1}{3}\left[2 \ln (x+3)+\ln x-\ln \left(x^{2}-1\right)\right]$$.

Step-by-Step Solution

Answer

The condensed form of the expression \( \frac{1}{3}\left[2 \ln (x+3)+\ln x-\ln\left(x^{2}-1\right)\right] \) is \( \ln \left( ((x^2+6x+9) ^{1/3}) / (x^{2}-1) \right) \)

Step 1: Apply the property of logs

The first step is to change each coefficient into an exponent in the logarithmic argument according to the formula \(a\ln b = \ln b^a\). In the provided expression prompt, there are multiple terms with coefficients, thus treating each term separately: \( \frac{1}{3} \cdot 2\ln(x+3) = \ln (x+3)^{2/3}\), and \( \frac{1}{3} \cdot \ln x = \ln x^{1/3}\) . There's no coefficient in \( \ln(x^{2}-1) \) .

Step 2: Apply Sum and Difference of Logs rule

After this transformation, the expression turns into \(\ln (x+3)^{2/3} + \ln x^{1/3} - \ln (x^{2}-1)\). This is a sum and difference of logs, and using the rule that \(\ln a + \ln b = \ln (ab), and \ln a - \ln b = \ln (a/b)\), we get \(\ln \left( (x+3)^{2/3} \cdot x^{1/3}\right) / (x^{2}-1)\)

Step 3: Calculate expression

Now we simplify \((x+3)^{2/3} \cdot x^{1/3}\): remember when you multiple two terms with same base then powers are added, \( (x+3)^{2/3} \cdot x^{1/3} = (x+3)^{2/3} \cdot (x)^{1/3} = (x^2+6x+9) ^{1/3} \). The final simplified expression becomes \( \ln \left( ((x^2+6x+9) ^{1/3}) / (x^{2}-1) \right) \)

Key Concepts

Condensing Logarithms

Condensing logarithms is an essential skill in managing complex logarithmic expressions. This process involves rewriting a long sum or difference of logarithms into a single logarithmic expression. Let's break down how condensing works.

When you have a logarithm like \(rac{1}{3}[2 \ln(x+3) + \ln x - \ln(x^{2}-1)]\), you start by manipulating each log term separately. This includes converting coefficients into exponents, which is crucial for the condensing process.

For instance, \(2 \ln(x+3)\) can be rewritten as \(\ln((x+3)^2)\) as per the property \(a\ln b = \ln b^a\). This makes it easier to combine these into a single log expression.

In the end, you should aim to express the original statement as \(\ln\) with one set of parentheses that includes all terms. This creates a streamlined expression that is easier to interpret and solve.

Exponents in Logarithms

Understanding how to deal with exponents in logarithms is fundamental to simplifying and manipulating logarithmic expressions.

The property \(a\ln b = \ln b^a\) allows coefficients in front of log expressions to be converted into exponents. For example, \(\frac{1}{3} \cdot 2 \ln(x+3)\) transforms into \(\ln((x+3)^{2/3})\).

Why do we do this? It's because exponents allow us to align multiple logarithmic terms for easier multiplication or division within the logarithmic rules.

By consistently applying this rule, you can simplify expressions significantly, which aids in solving equations or analyzing algebraic expressions involving logs.

Sum and Difference of Logarithms

One of the critical properties of logarithms is related to how we can sum or subtract them. This involves using the logarithmic identities:

• \(\ln a + \ln b = \ln(ab)\)
• \(\ln a - \ln b = \ln(\frac{a}{b})\)

When simplifying \(\ln (x+3)^{2/3} + \ln x^{1/3} - \ln (x^{2}-1)\), you apply these properties. First, combine the addition inside the logarithms using multiplication: \(\ln((x+3)^{2/3} \cdot x^{1/3})\).

Then, use the subtraction property to divide: \(\ln(\frac{(x+3)^{2/3} \cdot x^{1/3}}{x^{2}-1})\).

These rules enable you to condense multiple logarithmic terms into one, making problems easier to manage and solve efficiently.