Question

Write the following using logarithms: (a) \(32=2^{5}\) (b) \(125=5^{3}\) (c) \(243=3^{5}\) (d) \(4^{3}=64\) (e) \(6^{2}=36\)

Short answer

Question: Rewrite the following equations using logarithms: (a) 32=2^5 (b) 125=5^3 (c) 243=3^5 (d) 4^3=64 (e) 6^2=36 Answer: (a) \(log_2(32) = 5\) (b) \(log_5(125) = 3\) (c) \(log_3(243) = 5\) (d) \(log_4(64) = 3\) (e) \(log_6(36) = 2\)

Step-by-step solution

(a) Rewrite 32=2^5 using logarithms.

Using the logarithm formula \(log_b(x)=n\), we can rewrite the equation by taking logarithm base 2 for both sides, which will give us \(log_2(32) = log_2(2^5)\). Since \(log_b(b^n) = n\), the equation simplifies to \(log_2(32) = 5\).

(b) Rewrite 125=5^3 using logarithms.

Applying the same \(log_b(x)=n\) formula, we can rewrite the equation by taking logarithm base 5 for both sides, which will give us \(log_5(125) = log_5(5^3)\). Since \(log_b(b^n) = n\), the equation simplifies to \(log_5(125) = 3\).

(c) Rewrite 243=3^5 using logarithms.

Using the \(log_b(x)=n\) formula, we can rewrite the equation by taking logarithm base 3 for both sides, which will give us \(log_3(243) = log_3(3^5)\). Since \(log_b(b^n) = n\), the equation simplifies to \(log_3(243) = 5\).

(d) Rewrite 4^3=64 using logarithms.

Using the \(log_b(x)=n\) formula, we can rewrite the equation by taking logarithm base 4 for both sides, which will give us \(log_4(64) = log_4(4^3)\). Since \(log_b(b^n) = n\), the equation simplifies to \(log_4(64) = 3\).

(e) Rewrite 6^2=36 using logarithms.

Using the \(log_b(x)=n\) formula, we can rewrite the equation by taking logarithm base 6 for both sides, which will give us \(log_6(36) = log_6(6^2)\). Since \(log_b(b^n) = n\), the equation simplifies to \(log_6(36) = 2\).

Key concepts

Logarithmic Equations

Logarithmic equations are a type of algebraic equation in which the variable is contained within a logarithm. Understanding these equations is crucial for solving various problems in mathematics, especially those associated with exponential growth and decay.

To solve a logarithmic equation, one often starts by applying the definition of a logarithm. For instance, the equation \( \log_b(x) = n \) means \( b^n = x \), where \( b \) is the base of the logarithm, \( x \) is the argument, and \( n \) is the result or the 'power' to which the base \( b \) must be raised to obtain \( x \).

By understanding and manipulating these types of equations, students can solve complex problems involving exponential and logarithmic functions. In the provided exercise examples, we see this process unfold by converting exponential expressions into their equivalent logarithmic forms.

For instance, in solving the exercise part (a) \( 32=2^{5} \), we used the logarithmic equation \( \log_2(32) = 5 \), highlighting the basic principle that the exponent in the original expression becomes the result in the logarithmic form. Such translations from exponential to logarithmic expressions are fundamental in understanding the behavior of logarithms.

Logarithm Properties

Logarithm properties are like the Swiss Army knife for mathematicians - they provide a versatile set of tools that enable simplifying and computing logarithms effectively. Some of the essential properties include:

• The Product Rule: \( \log_b(x \cdot y) = \log_b(x) + \log_b(y) \), which implies that the logarithm of a product is the sum of the logarithms.
• The Quotient Rule: \( \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y) \), indicating that the logarithm of a quotient is the difference of the logarithms.
• The Power Rule: \( \log_b(x^n) = n \cdot \log_b(x) \), which allows us to take an exponent and move it to the front of the logarithm.
• The Change of Base Formula: \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \), where \( k \) is a new base, facilitating calculations with bases other than \( b \).

When working through logarithmic equations, these properties are often used in tandem to transform and solve for the unknown variable. They also help us understand why logarithms are inversely related to exponentiation - they essentially 'unpack' the exponent to reveal the power at which the base must be raised.

Exponential to Logarithmic Form

The ability to switch between exponential and logarithmic forms is a foundational skill in mathematics. This conversion is based on the intrinsic relationship between exponentiation and logarithms: logarithms answer the question 'to what power do we need to raise this base to get this number?'

An intuitive way to think about this connection is to consider that if you have an equation in the form of \( b^n = x \), you can translate it to its logarithmic counterpart \( \log_b(x) = n \). Here, \( b \) still represents the base, \( n \) is the exponent in the original expression, and \( x \) is the result of raising \( b \) to the power of \( n \).

This concept is demonstrated in the textbook solution where, for example, the exponential equation \( 4^3 = 64 \) is rephrased as \( \log_4(64) = 3 \). Through practice and application of this conversion process, students build a deeper understanding of the nature of exponentials and logarithms as two sides of the same mathematical coin.