Problem 11 - Calculus and its applications

Question

Solve for \(x\). $$ \log _{x} 32=5 $$

Step-by-Step Solution

Answer

The value of \( x \) is 2.

Step 1: Understand the Equation

The given equation is \( \log_{x} 32 = 5 \). This means that \( x \) is the base of the logarithm, and when \( x \) is raised to the power of 5, it equals 32.

Step 2: Convert the Logarithmic Equation to an Exponential Form

Logarithmic equations can be rewritten in exponential form. The equation \( \log_{x} 32 = 5 \) is equivalent to \( x^5 = 32 \). This is because the base \( x \) raised to the power 5 equals 32.

Step 3: Solve the Exponential Equation

We need to solve for \( x \) in the equation \( x^5 = 32 \). To find \( x \), take the fifth root of both sides of the equation: \( x = 32^{1/5} \).

Step 4: Calculate the Fifth Root

Evaluate \( 32^{1/5} \). Since \( 2^5 = 32 \), it follows that \( x = 2 \).

Key Concepts

Exponential Form

Converting a logarithmic equation into exponential form is a nifty trick used to make complicated problems simpler to solve. When you're working with logarithms, the core idea is about the relationship between different numbers as powers. In the given equation, \( \log_{x} 32 = 5 \), we want to express it in terms of exponents. This change takes advantage of the definition of logarithms. The equation says that \( x \) is the base, and when you raise \( x \) to the power of 5, you get 32. By rewriting it as \( x^5 = 32 \), it converts the logarithmic equation into something more familiar and easier to visualize. Using exponential form allows us to focus on solving for the base, \( x \), by understanding what power is required to result in 32.

Solving Equations

Solving the equation \( x^5 = 32 \) involves isolating \( x \). Once a logarithmic equation is written in its exponential form, we can leverage straightforward algebraic techniques to find its solution. Here, we aim to find \( x \) such that \( x^5 \) equals 32. The trick is to take the fifth root of both sides of the equation. This operation effectively "undoes" the exponent and isolates \( x \) on one side of the equation. By calculating the fifth root, we want \( x \) raised by the fifth power to return us to 32. Simplifying, \( x = 32^{1/5} \), and evaluating it reveals that \( x = 2 \) because \( 2^5 \) comfortably results in 32. Thus, through the fifth root, we've found \( x \), solving the equation effectively.

Logarithmic Functions

Logarithmic functions are unique in the sense that they are the inverse of exponential functions. Understanding these functions is key when you need to work with different forms of equations. If you think of exponentiation as a process of repeated multiplication, logarithms do the reverse—they ask how many times you need to multiply the base to reach a specific number.
In \( \log_{x} 32 = 5 \), the base \( x \) raised to the power of 5 equals 32. When converting this to exponential form as \( x^5 = 32 \), you reveal the underlying simplicity of logarithms: they disentangle the base and exponent to determine this very relationship between them. This makes logarithms particularly invaluable in fields requiring a deep understanding of growth processes and scaling, such as finance, sciences, and engineering.

• Inverse Nature: They 'undo' exponentiation.
• Applications: Essential in calculating growth rates and scaling equations.

Logarithmic functions, therefore, span a crucial part of mathematical problem-solving, offering powerful tools for both theoretical and practical applications.