Problem 33 - College Algebra with Corequisite Support

Question

For the following exercises, solve for \(x\) by converting the logarithmic equation to exponential form. \(\log _{6}(x)=-3\)

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{216} \).

Step 1: Understanding the Logarithmic Equation

The given equation is in the form of a logarithm: \( \log_{6}(x) = -3 \). This means we are looking for a value of \( x \) such that when 6 is raised to the power of \( -3 \), we get \( x \).

Step 2: Convert to Exponential Form

To convert the equation from logarithmic to exponential form, use the relationship: \( \log_b(a) = c \) can be rewritten as \( b^c = a \). Applying this to \( \log_{6}(x) = -3 \), we get \( 6^{-3} = x \).

Step 3: Calculate the Exponential Value

Now, calculate \( 6^{-3} \). Recall that raising a number to a negative exponent means taking the reciprocal of that number raised to the positive exponent. Thus, \( 6^{-3} = \frac{1}{6^3} \).

Step 4: Compute \( 6^3 \)

Calculate \( 6^3 \): \( 6 \times 6 \times 6 = 216 \).

Step 5: Final Calculation

Thus, \( 6^{-3} = \frac{1}{216} \). Therefore, \( x = \frac{1}{216} \).

Key Concepts

Logarithmic to Exponential Conversion

In mathematics, a logarithmic equation, such as \( \log_{b}(a) = c \), can be transformed into an exponential equation using the conversion principle that relates these expressions. Logarithmic and exponential forms are two sides of the same coin.Consider the form \( \log_{b}(a) = c \). This equation states that if the base \( b \) is raised to the power of \( c \), it will result in \( a \). Thus, it can be rewritten in its exponential form as \( b^c = a \). This conversion is incredibly useful when solving equations where \( x \) is the argument of the logarithm.When you see a logarithmic equation, think of it as an instruction: "What power do I need to raise \( b \) to get \( a \)?" This understanding helps one move seamlessly from a logarithmic form to an exponential one, making it easier to solve for unknown values. In the example \( \log_{6}(x) = -3 \), we apply this conversion to find that \( 6^{-3} = x \), which simplifies the process by maintaining the fundamental relationship between logarithms and exponents.

Negative Exponents

Negative exponents may seem a bit mysterious at first, but they follow a straightforward rule: they turn powers into fractions.When a number is raised to a negative exponent, it implies the reciprocal of the number raised to the absolute value of that exponent. This can be expressed as:

• \( a^{-n} = \frac{1}{a^n} \)

This means that instead of multiplying \( a \) by itself \( n \) times as in a positive exponent, you will divide 1 by \( a \) raised to that same \( n \).Applying this to our problem with the equation \( 6^{-3} \), it follows that:

• \( 6^{-3} = \frac{1}{6^3} \)
• \( 6^3 \) is computed as \( 6 \times 6 \times 6 = 216 \).
• Thus, \( 6^{-3} = \frac{1}{216} \).

Understanding negative exponents allows you to handle problems where large numbers become very small through division, making it an invaluable tool when working with exponential equations.

Solving Logarithmic Equations

Solving logarithmic equations often entails converting the logarithm into a form that reveals the unknown variable. The key to solving these types of equations lies in converting from a logarithmic to an exponential form, as discussed earlier.For instance, given \( \log_{b}(x) = c \), the target is to express this equation in terms of an exponential function: \( b^c = x \). This swap simplifies extracting the unknown value because exponential equations are more straightforward to handle algebraically.To solve the equation \( \log_{6}(x) = -3 \):

• First, convert the logarithm to its exponential form, which results in \( x = 6^{-3} \).
• Next, handle the negative exponent by computing \( 6^{-3} \) as \( \frac{1}{6^3} \).
• Calculate \( 6^3 \) which is \( 216 \), thereby \( x = \frac{1}{216} \).

In doing so, solving the equation becomes a matter of simple arithmetic, helping you find the solution efficiently. By tackling the logarithmic equation through these steps, one experiences the power of converting and solving logarithmic equations with clarity and confidence.