Problem 273 - Calculus Volume 1

Question

Use properties of logarithms to write the expressions as a sum, difference, and/or product of logarithms. \(\log _{5} \sqrt{125 x y^{3}}\)

Step-by-Step Solution

Answer

\( \frac{3}{2} + \frac{1}{2} \log_5(x) + \frac{3}{2} \log_5(y) \)

Step 1: Express the square root as an exponent

Start by rewriting the expression inside the logarithm to handle the square root. Recall that a square root can be expressed as a power of one-half. Thus, rewrite \( \sqrt{125xy^3} \) as \( (125xy^3)^{\frac{1}{2}} \).

Step 2: Apply the power rule of logarithms

Use the power rule of logarithms, which states that \( \log_b(a^c) = c \cdot \log_b(a) \). Applying this rule gives us \( \frac{1}{2} \cdot \log_5 (125xy^3) \).

Step 3: Break down the product inside the logarithm

Next, use the property of logarithms that states \( \log_b(abc) = \log_b(a) + \log_b(b) + \log_b(c) \). Apply this to separate \( \log_5 (125xy^3) \) into \( \log_5 (125) + \log_5 (x) + \log_5 (y^3) \).

Step 4: Apply the power rule to the last term

There's another power inside the logarithm: \( y^3 \). Use the power rule again to express \( \log_5 (y^3) \) as \( 3 \cdot \log_5 (y) \).

Step 5: Combine all terms

Now combine everything back together: \( \frac{1}{2} (\log_5 (125) + \log_5 (x) + 3 \cdot \log_5 (y)) \).

Step 6: Simplify where possible

Finally, determine if any expressions can be simplified further. Note where \( \log_5 (125) \) can be simplified because \( 125 = 5^3 \). Therefore, \( \log_5 (125) = 3 \cdot \log_5 (5) = 3 \cdot 1 = 3 \). Substitute back: \( \frac{1}{2} (3 + \log_5 (x) + 3 \cdot \log_5 (y)) \). Distribute the \( \frac{1}{2} \) to get \( \frac{3}{2} + \frac{1}{2} \cdot \log_5 (x) + \frac{3}{2} \cdot \log_5 (y) \).

Key Concepts

Power Rule of Logarithms

The power rule of logarithms is a handy tool when working with exponential expressions inside logarithms. It allows us to move the exponent out of the logarithm, simplifying the expression significantly.
This rule states that \( \log_b(a^c) = c \cdot \log_b(a) \). In essence, you can "bring down" the exponent to the front of the logarithm as a coefficient.

• For example, if you have \( \log_5(125xy^3) \), and you want to rewrite \( y^3 \), you apply this rule to get \( 3 \cdot \log_5(y) \).
• This transformation makes it much easier to deal with expressions involving powers or roots inside logarithms.

Understanding the power rule is crucial because it gives you a method to handle expressions that initially might look difficult to manipulate. Always look for opportunities to apply this rule when you have exponents lurking inside a logarithm.

Product Rule of Logarithms

The product rule of logarithms is another vital property that breaks down the product inside a logarithm into separate logarithmic expressions. It transforms multiplication into addition, allowing you to spread out complex terms.
The rule is expressed as \( \log_b(abc) = \log_b(a) + \log_b(b) + \log_b(c) \). Essentially, you split the single logarithm of a product into the sum of individual logarithms.

• Applying this rule to \( \log_5(125xy^3) \) would give you \( \log_5(125) + \log_5(x) + \log_5(y^3) \).
• This breaking down simplifies not just understanding the expression but also in performing further simplifications.

The product rule is especially useful in situations where combining terms is necessary for easier handling of more complex expressions.

Simplification of Logarithms

Simplification of logarithmic expressions involves applying various properties of logarithms to make the expression as concise and manageable as possible. After applying rules such as the power rule and the product rule, you should always look for further simplifications.For instance, when working with \( \log_5(125) \), recognize that 125 is equal to \( 5^3 \). Therefore, \( \log_5(125) \) simplifies to \( 3 \cdot \log_5(5) \), which is simply 3, since \( \log_5(5) = 1 \).

• After initial breakdown using logarithmic rules, always combine constants, like \( \frac{3}{2} \) in the given solution.
• Simplification often includes distributing coefficients across terms, like from \( \frac{1}{2} (3 + \log_5 (x) + 3 \cdot \log_5 (y)) \) to final form \( \frac{3}{2} + \frac{1}{2} \cdot \log_5 (x) + \frac{3}{2} \cdot \log_5 (y) \).

Consistent practice in simplifying can demystify logarithms, making them more intuitive and easier to work with in various mathematical contexts.