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Chapter 5: Problem 69

Use the properties of logarithms to rewrite each logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{5} \frac{5 \sqrt{7}}{3 m}$$

Short Answer

Expert verified
\( 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \)

Step by step solution

01

Apply the Quotient Rule

The Quotient Rule of logarithms states that \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \). We apply this rule to \( \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) \): \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = \log_5 (5 \sqrt{7}) - \log_5 (3m) \]
02

Apply the Product Rule

The Product Rule of logarithms is \( \log_b (xy) = \log_b x + \log_b y \). We apply it to \( \log_5 (5 \sqrt{7}) \): \[ \log_5 (5 \sqrt{7}) = \log_5 5 + \log_5 \sqrt{7} \]
03

Simplify using the logarithm of a power

The logarithm \( \log_5 5 = 1 \) because any logarithm of its own base is 1. Also, rewrite \( \log_5 \sqrt{7} \) using the power rule: \[ \sqrt{7} = 7^{1/2}, \text{ so } \log_5 \sqrt{7} = \log_5 (7^{1/2}) = \frac{1}{2} \log_5 7 \] Thus, \[ \log_5 (5 \sqrt{7}) = 1 + \frac{1}{2} \log_5 7 \]
04

Apply the Product Rule to the denominator

Apply the Product Rule to the denominator term \( \log_5 (3m) \), breaking it into two separate logarithms: \[ \log_5 (3m) = \log_5 3 + \log_5 m \]
05

Combine all parts

Combine the results from previous steps. Use the equation from Step 1 and substitute the results from Step 2 and Step 4: \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = (1 + \frac{1}{2} \log_5 7) - (\log_5 3 + \log_5 m) \] Simplifying gives: \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \]
06

Final expression

The final expression, using the properties of logarithms, is: \[ 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
The quotient rule in logarithms helps us decompose a logarithm of a fraction into a subtraction of two logs. It's a valuable tool when you're working with expressions that involve division inside a logarithm. Here's how it works: if you have a logarithm of a fraction like \( \log_b \left( \frac{x}{y} \right) \), you can split it into two parts: \( \log_b x - \log_b y \). In practice, this means you can
  • change a complex division problem into simpler subtraction problems,
  • making it easier to solve or simplify the expression efficiently.
For example, with our exercise, \( \log_{5} \left( \frac{5 \sqrt{7}}{3m} \right) \) becomes \( \log_{5} (5\sqrt{7}) - \log_{5} (3m) \) using the quotient rule. This step simplifies the problem by separating the numerator and the denominator.
Product Rule
The product rule of logarithms is a handy property for breaking down logs of products into sums of logs. The rule states that for any logarithm \( \log_b (xy) = \log_b x + \log_b y \). This is useful because
  • we can break down a complex multiplication inside a logarithm into simpler addition expressions,
  • facilitating the step-by-step simplification or calculation process.
Consider the part of our exercise where we deal with \( \log_{5} (5\sqrt{7}) \). By applying the product rule, we rewrite it as \( \log_{5} 5 + \log_{5} \sqrt{7} \). This step makes each individual term easier to handle.
Logarithm Properties
Logarithm properties encompass various rules that help in simplifying and solving logarithmic expressions. Some crucial properties include:
  • The identity \( \log_b b = 1 \), where any log that has the same base and argument equals 1.
  • Using log transformations for powers and radicals which are explained by the power rule.
In our example, applying these properties reveals that \( \log_{5} 5 = 1 \), which simplifies part of the expression straight away. Utilizing these core properties can significantly streamline the handling of logs.
Logarithm of a Power
Logarithms of powers can be simplified using the power rule. This rule says \( \log_b (x^a) = a \cdot \log_b x \), which means you can take the exponent and place it as a multiplier in front of the logarithm.For instance,
  • if you have a square root like \( \sqrt{7} \), this is the same as \( 7^{1/2} \).
  • Applying the power rule, \( \log_5 \sqrt{7} \) becomes \( \frac{1}{2} \log_5 7 \).
This step is important as it transforms complex expressions into computationally friendlier terms. Simplifying \( \log_5 (5 \sqrt{7}) \) with \( 1 + \frac{1}{2} \log_5 7 \) shows how breaking down a power log aids in simplification.

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Most popular questions from this chapter

Sound Intensity Use the formula $$d=10 \log \frac{I}{I_{0}}$$ to estimate the average decibel level for each sound with the given intensity \(I .\) For comparison, conversational speech has a sound level of about 60 decibels. (a) Jackhammer: \(31,620,000,000 I_{0}\). (b) iPhone 5 speakers: \(10^{1 /} l_{0}\). (c) Rock singer screaming into microphone: \(10^{14} I_{0}\).

Use a graphing calculator to solve each equation. Give solutions to the nearest hundredth. \(\log _{10} x=x-2\)

If interest is compounded continuously and the interest rate is tripled, what effect will this have on the time required for an investment to double?

Tom Tupper wants to buy a \(\$ 30,000\) car. He has saved \(\$ 27,000 .\) Find the number of years (to the nearest tenth) it will take for his \(\$ 27,000\) to grow to \(\$ 30,000\) at \(2.3 \%\) interest compounded quarterly.

Assume that \(f(x)=a^{x},\) where \(a>1\) If \(a=e,\) what is an equation for \(y=f^{-1}(x) ?\) (You need not solve for \(y .\) )
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