Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, solve for \(x\) by converting the logarithmic equation to exponential form. \(\log _{2}(x)=-3\)

Step-by-Step Solution

Answer

\( x = \frac{1}{8} \)

Step 1: Understand the relationship between logarithms and exponents

When you have a logarithmic equation like \( \log_b(a) = c \), it can be converted to its exponential form, which is \( b^c = a \). This is because logarithms are the inverse operation of exponentiation.

Step 2: Identify the base, the argument, and the result

In the equation \( \log_{2}(x) = -3 \), identify the base as 2, the argument as \( x \), and the result as \( -3 \). This will help in converting the equation into its exponential form.

Step 3: Convert the logarithmic form to exponential form

Use the relationship from Step 1 to rewrite the equation: \( \log_{2}(x) = -3 \) becomes \( 2^{-3} = x \). Here, you are saying that 2 raised to the power of \( -3 \) equals \( x \).

Step 4: Calculate the exponential expression

Now compute \( 2^{-3} \). Recall that a negative exponent means taking the reciprocal: \( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \).

Step 5: Confirm the solution

Check the solution by substituting back into the original equation to verify: Compute \( \log_{2}(\frac{1}{8}) \): since \( 2^3 = 8 \), and \( 2^{-3} = \frac{1}{8} \), the equation holds true.

Key Concepts

Exponential Form

To solve logarithmic equations like \( \log_b(a) = c \), we convert them into a form that is more straightforward to handle, known as exponential form. The equation \( \log_b(a) = c \) translates to \( b^c = a \). This transformation happens because logarithms are essentially the inverse of exponential functions. They unravel the exponentiation process, providing a clearer path to find the unknown variable.
Understanding this relationship is crucial, as it enables us to handle equations and problems where the log expression might seem complex at first glance. When you convert to exponential form, it translates the mystery of the logarithm into a power equation that can be more easily manipulated and solved.

Logarithms and Exponents

The interplay between logarithms and exponents is a fundamental concept in mathematics. Logarithms, often perceived as complicated, mirror the process of exponential functions. When you encounter an equation in the form \( \log_b(x) = y \), what this tells you is that you need to find the power \( y \) that the base \( b \) must be raised to in order to yield \( x \).
- **Logarithmic Notation:** \( \log_b(a) = c \) means the exponent \( c \) is what you need to raise \( b \) to get \( a \).- **Exponential Notation:** Once converted, the form becomes \( b^c = a \), showing the numeric relationship more explicitly.This equating of concepts shows how solving logarithmic problems often involves stepping back into a basic understanding of powers and exponents.

Negative Exponents

Negative exponents often introduce confusion, but they are simpler than they seem. A negative exponent indicates that we take the reciprocal of the base raised to the positive of that exponent. For example, \( 2^{-3} \) is computed as the reciprocal of \( 2^3 \). Mathematically, this can be expressed as:- \( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \).This concept is crucial when solving equations that involve exponents. In the context of converting logarithms to exponential form, realizing that a negative exponent means dividing rather than multiplying can simplify seemingly complex equations.
Understanding negative exponents also aids in seeing how small a value becomes as the negative exponent increases, indicating a diminishing fraction or decimal when computed.