Problem 35 - Precalculus Mathematics for Calculus

Question

Use the definition of the logarithmic function to find \(x .\) (a) \(\log _{x} 16=4\) (b) \(\log _{x} 8=\frac{3}{2}\)

Step-by-Step Solution

Answer

(a) \( x = 2 \), (b) \( x = 4 \)

Step 1: Understanding the Logarithmic Definition

Recall the definition of a logarithmic function: if \( \log_b a = c \), then \( b^c = a \). This means a logarithm is an exponent. We will use this definition to solve for \( x \) in both parts of the problem.

Step 2: Solve Part (a): Convert the Logarithmic Equation to an Exponent Equation

For part (a), we need to solve \( \log_x 16 = 4 \). According to the definition, this converts to the exponential equation \( x^4 = 16 \).

Step 3: Simplify the Equation for Part (a)

To solve \( x^4 = 16 \), take the fourth root of both sides. This gives \( x = \sqrt[4]{16} \). Since \( 16 = 2^4 \), we get \( x = 2 \).

Step 4: Solve Part (b): Convert the Logarithmic Equation to an Exponent Equation

For part (b), we need to solve \( \log_x 8 = \frac{3}{2} \). Using the logarithm definition, this converts to the exponential equation \( x^{3/2} = 8 \).

Step 5: Simplify the Equation for Part (b)

To solve \( x^{3/2} = 8 \), raise both sides to the power of \( \frac{2}{3} \) to isolate \( x \). Calculating this gives \( x = 8^{\frac{2}{3}} \).

Step 6: Calculate \( 8^{\frac{2}{3}} \)

Since \( 8 = 2^3 \), \( 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^{3 \times \frac{2}{3}} = 2^2 = 4 \). Thus, \( x = 4 \).

Key Concepts

Exponential Equations

Exponential equations are mathematical expressions where variables appear as exponents. They are crucial in solving problems where growth or decay is based on a constant multiplier. An exponential equation has the form \( b^c = a \), where \( b \) is the base, \( c \) is the exponent, and \( a \) is the result. When solving exponential equations, the goal is often to isolate the base variable to determine its value. For example, to solve the equation \( x^4 = 16 \), we recognize this as a classical exponential problem. By taking the fourth root of both sides, we isolate \( x \), resulting in \( x = \sqrt[4]{16} = 2 \). Similarly, consider solving \( x^{3/2} = 8 \). This requires manipulating the equation by raising both sides to the reciprocal power \( \frac{2}{3} \) as in \( (x^{3/2})^{\frac{2}{3}} = 8^{\frac{2}{3}} \), which simplifies to finding the cube root first and then squaring the result.

Logarithmic Definition

The logarithmic function is the inverse operation of exponential functions. This is fundamentally expressed as \( \log_b a = c \), which means the base \( b \) raised to the exponent \( c \) equals \( a \). Logarithms can be thought of as solving for the exponent in exponential equations. They are incredibly useful in cases where the value of the exponent isn't immediately clear. For problems like \( \log_x 16 = 4 \), using the logarithmic definition, we convert and interpret it as the exponential equation \( x^4 = 16 \). This identifies the unknown \( x \) as something we need to solve for using knowledge about powers and roots. In \( \log_x 8 = \frac{3}{2} \), the definition becomes \( x^{3/2} = 8 \). Recognizing this form assists us in calculating the base \( x \) by reversing exponential operations, such as taking roots or lifting both sides to a power.

Roots of Numbers

Roots of numbers are operations that revert powers and are significant in solving equations involving exponents. The "root" operation answers the question "which number, multiplied by itself several times, equals a given number?" For instance, the fourth root \( \sqrt[4]{16} \) gives the number \( 2 \) because \( 2^4 = 16 \). When dealing with fractional powers, as in the equation \( x^{3/2} = 8 \), we use roots creatively:

• The fraction \( \frac{3}{2} \) indicates both a square and a cube process. First, the cube root \( 8^{1/3} = 2 \) is calculated.
• This result is then squared to find \( 2^2 = 4 \).

Understanding how to calculate roots and work with fractional exponents equips us to transform complex equations into simple arithmetic.