Problem 70
Question
Use the properties of logarithms to rewrite each logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{2} \frac{2 \sqrt{3}}{5 p}$$
Step-by-Step Solution
Verified Answer
Rewrite as: \( \log_2 \frac{2 \sqrt{3}}{5p} = 1 + \frac{1}{2} \log_2 3 - \log_2 5 - \log_2 p \).
1Step 1: Apply the Quotient Rule
The quotient rule of logarithms states that \( \log_a \frac{b}{c} = \log_a b - \log_a c \). Applying this rule to the expression \( \log_2 \frac{2 \sqrt{3}}{5p} \), we get: \( \log_2 (2 \sqrt{3}) - \log_2 (5p) \).
2Step 2: Apply the Product Rule
The product rule of logarithms states that \( \log_a (b \, c) = \log_a b + \log_a c \). For \( \log_2 (2 \sqrt{3}) \), we can use this rule to get: \( \log_2 2 + \log_2 \sqrt{3} \).
3Step 3: Simplify Using Known Log Values
We know that \( \log_2 2 = 1 \) because 2 is the base of the logarithm. Therefore, \( \log_2 2 + \log_2 \sqrt{3} = 1 + \log_2 \sqrt{3} \).
4Step 4: Rewrite the Square Root Using Exponent
\( \sqrt{3} \) can be rewritten as \( 3^{1/2} \). Therefore, \( \log_2 \sqrt{3} = \log_2 (3^{1/2}) = \frac{1}{2} \log_2 3 \) using the power rule.
5Step 5: Combine the Expression
Substitute back the values from Steps 2 and 4 into the expression. We have \( 1 + \frac{1}{2} \log_2 3 \) from the first term and subtract \( \log_2 5 + \log_2 p \) from the second term. This results in the final expression: \( 1 + \frac{1}{2} \log_2 3 - \log_2 5 - \log_2 p \).
Key Concepts
Quotient RuleProduct RulePower RuleLogarithm Simplification
Quotient Rule
The quotient rule in logarithms is an essential property that helps to simplify expressions involving the division of numbers or variables. According to the quotient rule, when you have the logarithm of a fraction, such as \( \log_a \frac{b}{c} \), it can be rewritten as the difference of two separate logarithms: \( \log_a b - \log_a c \).
This rule is particularly helpful because it breaks down complex expressions into simpler components.
For example, in the expression \( \log_2 \frac{2 \sqrt{3}}{5p} \), the quotient rule allows us to express it as \( \log_2 (2 \sqrt{3}) - \log_2 (5p) \).
This transformation turns one challenging fraction into two straightforward terms, setting the stage for further simplification.
This rule is particularly helpful because it breaks down complex expressions into simpler components.
For example, in the expression \( \log_2 \frac{2 \sqrt{3}}{5p} \), the quotient rule allows us to express it as \( \log_2 (2 \sqrt{3}) - \log_2 (5p) \).
This transformation turns one challenging fraction into two straightforward terms, setting the stage for further simplification.
Product Rule
The product rule of logarithms is another important tool for simplifying logarithmic expressions. This rule tells us that the logarithm of a product can be written as the sum of the logarithms of its factors. Mathematically, it appears as \( \log_a (b \, c) = \log_a b + \log_a c \).
The product rule essentially transforms a single compounded term into two simpler parts, allowing for easier manipulation and understanding.
In this exercise, the expression \( \log_2 (2 \sqrt{3}) \) can be decomposed using the product rule into \( \log_2 2 + \log_2 \sqrt{3} \).
This breakdown simplifies the logarithmic process, making each term easier to work with individually.
The product rule essentially transforms a single compounded term into two simpler parts, allowing for easier manipulation and understanding.
In this exercise, the expression \( \log_2 (2 \sqrt{3}) \) can be decomposed using the product rule into \( \log_2 2 + \log_2 \sqrt{3} \).
This breakdown simplifies the logarithmic process, making each term easier to work with individually.
Power Rule
The power rule of logarithms provides a way to deal with expressions where numbers or variables are raised to a power.
This rule states that \( \log_a (b^c) = c \cdot \log_a b \), allowing you to "bring down" the exponent as a coefficient in front of the logarithm.
Consider the portion of our exercise involving \( \log_2 \sqrt{3} \). Knowing that \( \sqrt{3} \) is equivalent to \( 3^{1/2} \), this can be rewritten using the power rule: \( \log_2 (3^{1/2}) = \frac{1}{2} \log_2 3 \).
By turning the exponent into a multiplier, it becomes easier to manage logarithmic expressions with exponents.
This rule states that \( \log_a (b^c) = c \cdot \log_a b \), allowing you to "bring down" the exponent as a coefficient in front of the logarithm.
Consider the portion of our exercise involving \( \log_2 \sqrt{3} \). Knowing that \( \sqrt{3} \) is equivalent to \( 3^{1/2} \), this can be rewritten using the power rule: \( \log_2 (3^{1/2}) = \frac{1}{2} \log_2 3 \).
By turning the exponent into a multiplier, it becomes easier to manage logarithmic expressions with exponents.
Logarithm Simplification
Simplifying logarithmic expressions often involves the application of the quotient, product, and power rules, as seen in the provided problem.
To reach a fully simplified outcome, each term is methodically broken down to its simplest components and then combined back into a coherent expression.
In our specific exercise, after applying all rules and simplifying known values such as \( \log_2 2 = 1 \), the expression was gradually transformed.
To reach a fully simplified outcome, each term is methodically broken down to its simplest components and then combined back into a coherent expression.
In our specific exercise, after applying all rules and simplifying known values such as \( \log_2 2 = 1 \), the expression was gradually transformed.
- First, the application of rules broke down the expression \( \log_2 \frac{2 \sqrt{3}}{5p} \).
- Next, known values and rules condensed it into \( 1 + \frac{1}{2} \log_2 3 - \log_2 5 - \log_2 p \).
Other exercises in this chapter
Problem 69
$$\text { Solve each formula for the indicated variable.}$$ $$r=p-k \ln t, \text { for } t$$
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Use the appropriate compound interest formula to find the amount that will be in each account, given the stated conditions. \(\$ 35,000\) invested at \(4.2 \%\)
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The given function \(f\) is one-to-one. Find \(f^{-1}(x)\). $$f(x)=\frac{4-x}{5 x}$$
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$$\text { Solve each formula for the indicated variable.}$$ $$p=a+\frac{k}{\ln x}, \text { for } x$$
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