Problem 29 - Algebra and Trigonometry

Question

Use the definition of the logarithmic function to find \(x\) $$ \begin{array}{ll}{\text { (a) } \log _{2} x=5} & {\text { (b) } \log _{2} 16=x}\end{array} $$

Step-by-Step Solution

Answer

(a) x = 32, (b) x = 4

Step 1: Understand the problem

We need to solve two logarithmic equations using the definition of logarithms. The logarithmic function \( \log_b a = c \) means \( b^c = a \).

Step 2: Use the definition for equation (a)

For equation (a), \( \log_{2} x = 5 \), we apply the definition: \( 2^5 = x \).

Step 3: Compute for equation (a)

Calculate \( 2^5 = 32 \). Therefore, \( x = 32 \) for the first part.

Step 4: Use the definition for equation (b)

For equation (b), \( \log_{2} 16 = x \), apply the definition: \( 2^x = 16 \).

Step 5: Compute for equation (b)

Observe that \( 2^4 = 16 \). Therefore, \( x = 4 \) for the second part.

Key Concepts

Logarithms

Logarithms are a fundamental concept in mathematics, often acting as the reverse of exponentiation. They help us solve equations where an unknown appears as an exponent. The logarithmic function is written as \( \log_b a = c \), which reads as "\(c\) is the power to which the base \(b\) must be raised to obtain \(a\)."
For example, if you have \( \log_{10} 100 = 2 \), it implies that \(10^2 = 100\). Logarithms are immensely useful in various fields like science and engineering, where exponential growth processes are common.
When dealing with logarithms, note these key properties:

• Product Rule: \( \log_b (mn) = \log_b m + \log_b n \)
• Quotient Rule: \( \log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n \)
• Power Rule: \( \log_b (m^n) = n \cdot \log_b m \)

Exponentiation

Exponentiation is the mathematical operation involving two numbers, the base and the exponent. It describes processes where numbers are multiplied by themselves a certain number of times, defined as \( b^n \), where \(b\) is the base and \(n\) is the exponent.
If \( n \) is a positive integer, then the expression \( b^n \) represents the multiplication of the base \(b\) exactly \( n \) times. For example, \( 2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32 \).

• Zero Exponent Rule: Any non-zero base \(b\) raised to the power of 0 is 1, i.e., \( b^0 = 1 \).
• Negative Exponent Rule: A negative exponent indicates the reciprocal, i.e., \( b^{-n} = \frac{1}{b^n} \).
• Fractional Exponents: Correspond to roots, such as \( b^{1/2} = \sqrt{b} \).

Solving Logarithmic Equations

To solve logarithmic equations, the key is to understand and manipulate the equation using logarithmic properties or by converting the logarithmic equation into its exponential form.
Consider the example equation \( \log_{b} x = a \). By transforming this into its exponential form, \( b^a = x \), you can solve for \(x\).
Let's solve a typical exercise to solidify this:

• For \( \log_{2} x = 5 \), convert to the exponential form to get \( 2^5 = x \), thus \( x = 32 \).
• For \( \log_{2} 16 = x \), again convert to the exponential form, \( 2^x = 16 \). Noticing that \( 2^4 = 16 \), we find \( x = 4 \).

This step-by-step conversion breaks down the problem and highlights how to rearrange equations efficiently. It demonstrates the power of converting logarithmic expressions to exponential form for easier computation.