Problem 47 - Intermediate Algebra

Question

Solve each equation. $$ \log \frac{1}{8} x=-2 $$

Step-by-Step Solution

Answer

The value of \( x \) is \( \frac{2}{25} \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log \left( \frac{1}{8} x \right) = -2 \). This is a logarithmic equation where we need to find the value of \( x \).

Step 2: Converting Logarithmic to Exponential Form

Recall that if \( \log_b A = C \), it means \( A = b^C \). Here, we assume the base of the logarithm is 10, so \( \frac{1}{8} x = 10^{-2} \).

Step 3: Calculate the Exponential

Calculate \( 10^{-2} \), which equals \( \frac{1}{100} \). Therefore, the equation becomes \( \frac{1}{8} x = \frac{1}{100} \).

Step 4: Solve for x

To isolate \( x \), multiply both sides of the equation by 8: \( x = \frac{1}{100} \times 8 \).

Step 5: Simplify the Expression

Simplify \( \frac{8}{100} \) by dividing the numerator and the denominator by 4: \( \frac{8}{100} = \frac{2}{25} \). Therefore, \( x = \frac{2}{25} \).

Key Concepts

Logarithmic Equations

Logarithmic equations are an integral part of intermediate algebra that involve logarithms, which are operations that tell us how many times a number, called the 'base', is multiplied by itself to reach another number.
Given an equation like \( \log \left( \frac{1}{8} x \right) = -2 \), we are working with a base-10 logarithm unless specified otherwise.

• The key principle here is understanding what a logarithm does. It helps to reverse the process of exponentiation.
• Expressing an equation in logarithmic form allows us to solve for unknowns where they function as exponents or multiplicative factors.
• This is crucial for equations where the unknown is inside the logarithm, as seen in your exercise.

By practicing these, you build intuition on transforming and manipulating such equations to isolate and solve for the unknown.

Exponential Form

Converting a logarithmic equation to exponential form is a vital skill that can simplify solving. The relationship between these forms is foundational in algebra.

• If you have \( \log_b A = C \), you can express it in exponential form as \( A = b^C \).
• This conversion switches the perspective from asking "what power gives me a number" (logarithmic) to "power a base to get a number" (exponential).
• For example, in the equation \( \log \left( \frac{1}{8} x \right) = -2 \), we assume a base of 10 to rewrite it as \( \frac{1}{8} x = 10^{-2} \).

Understanding this switch is crucial for progressing in algebra, especially when the function needs rearranging to solve easily.

Solving Equations

Once a logarithmic equation is in exponential form, solving becomes clearer. Solving involves algebraic manipulation to isolate the unknown, often requiring further simplification.

• Start by evaluating the exponential expression. In this exercise, \( 10^{-2} \) converts to \( \frac{1}{100} \).
• With the equation \( \frac{1}{8} x = \frac{1}{100} \), we can solve for \( x \) by multiplying to clear fractions.
• Multiplying both sides by 8 results in \( x = \frac{1}{100} \times 8 \), simplifying further by reducing the fraction.

This tactic is more straightforward once the equation is simplified, guiding towards neat resolution of the equation. It's important to pay attention to proper multiplication and reduction to prevent arithmetic errors.