Question

If \(\log _{2} 7=K,\) write an algebraic expression in terms of \(K\) for each of the following. a) \(\log _{2} 7^{6}\) b) \(\log _{2} 14\) c) \(\log _{2}(49 \times 4)\) d) \(\log _{2} \frac{\sqrt[5]{7}}{8}\)

Short answer

a) 6K, b) K + 1, c) 2K + 2, d) \frac{K}{5} - 3

Step-by-step solution

Understanding the Given

Given \(\text{log}_{2}(7) = K\). This is the starting point to derive the expressions for parts a, b, c, and d.

Solving Part (a) \(\log_{2} 7^{6}\)

Use the power rule of logarithms: \( \log_{b}(a^n) = n \, \log_{b}(a) \). Thus, \(\text{log}_{2}(7^{6}) = 6 \cdot \text{log}_{2}(7) \). Since \(\text{log}_{2}(7) = K\), it follows that \(\text{log}_{2}(7^{6}) = 6K\).

Solving Part (b) \(\log_{2}(14)\)

Use the product rule of logarithms: \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \). Here, \(14 = 7 \times 2 \), then \(\log_{2}(14) = \log_{2}(7 \times 2) = \log_{2}(7) + \log_{2}(2) \). Given \( \log_{2}(7) = K \) and knowing that \(\log_{2}(2) = 1 \), it follows that \(\log_{2}(14) = K + 1\).

Solving Part (c) \(\log_{2}(49 \times 4)\)

Use the product rule: \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \). Here, \(49 = 7^2 \) and \(4 = 2^2 \), so \(49 \times 4 = 7^2 \times 2^2 = (7 \times 2)^2 \). Thus, \(\log_{2}(49 \times 4) = \log_{2}( (7 \times 2)^2) = 2 \, \log_{2}(7 \times 2)\). Using the product rule: \(\log_{2}(7 \times 2) = \log_{2}(7) + \log_{2}(2) = K + 1\). Therefore, \(\log_{2}(49 \times 4) = 2(K + 1) = 2K + 2\).

Solving Part (d) \(\log_{2} \frac{\sqrt[5]{7}}{8} \)

Use the quotient rule of logarithms: \( \log_{b} \left( \frac{x}{y} \right) = \log_{b}(x) - \log_{b}(y) \). Here, \(\frac{\sqrt[5]{7}}{8} = 7^{1/5} \times 8^{-1} = 7^{1/5} \times (2^3)^{-1} = 7^{1/5} \times 2^{-3} \). Apply the product rule: \(\log_{2} (\sqrt[5]{7} \times 2^{-3}) = \log_{2}(7^{1/5}) + \log_{2}(2^{-3}) \). Using the power rule: \( \log_{2}(7^{1/5}) = \frac{1}{5} \log_{2}(7) = \frac{K}{5} \) and \( \log_{2}(2^{-3}) = -3 \log_{2}(2) = -3 \). So, \( \log_{2} \left( \frac{\sqrt[5]{7}}{8} \right) = \frac{K}{5} - 3 \).

Key concepts

logarithm properties

Logarithms have interesting and useful properties that help simplify complex expressions. Understanding these properties makes it easier to work with logarithmic equations and solve problems. Here are some key properties:

• Product Rule: \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \)
• Quotient Rule: \( \log_{b}\left( \frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y) \)
• Power Rule: \( \log_{b}(a^n) = n\log_{b}(a) \)
• Base Rule: \( \log_{b}(b) = 1 \)

These properties are helpful in breaking down logarithmic expressions for easier calculation.

product rule of logarithms

The product rule is an essential logarithm property. It states that the logarithm of a product is the sum of the logarithms of the individual factors. Formally, it is written as:

\( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \)

For example, when solving the problem \( \log_{2}(14) \), we identify that \( 14 = 7 \times 2 \). Using the product rule:

\( \log_{2}(14) = \log_{2}(7 \times 2) = \log_{2}(7) + \log_{2}(2) \).

Given that \( \log_{2}(7) = K \) and knowing that \( \log_{2}(2) = 1 \),
It follows that \( \log_{2}(14) = K + 1 \).

This rule simplifies multiplication inside a logarithm, making calculations more manageable.

quotient rule of logarithms

The quotient rule states that the logarithm of a quotient is the difference between the logarithms of the numerator and the denominator. Formally, it looks like this:

\( \log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y) \)

This rule is very useful in simplifying the division inside a logarithm. For example, when solving \( \log_{2}\left(\frac{\sqrt[5]{7}}{8}\right) \), we rewrite it using the quotient rule:

\( \log_{2}\left(\frac{\sqrt[5]{7}}{8}\right) = \log_{2}(\sqrt[5]{7}) - \log_{2}(8) \).

Next, use the power rule where applicable to break it down further. Understanding and applying the quotient rule makes logarithmic division much simpler and more intuitive.

power rule of logarithms

The power rule states that the logarithm of a power is the exponent multiplied by the logarithm of the base. Formally:

\( \log_{b}(a^n) = n\log_{b}(a) \)

This property helps in dealing easily with exponential values within logarithms. For instance, in solving \( \log_{2}(7^6) \), we use the power rule:

\( \log_{2}(7^6) = 6\log_{2}(7) \).

Given that \( \log_{2}(7) = K \), it follows that \( \log_{2}(7^6) = 6K \).

The power rule simplifies exponents within a logarithm, making it easier to manage and calculate logarithmic expressions
Understanding and using these rules effectively can greatly streamline your work with logarithms.