Problem 32 - College Algebra

Question

For the following exercises, use the definition of a logarithm to solve the equation. $$ -8 \log _{9}(x)=16 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{81} \).

Step 1: Isolate the Logarithm

Begin by isolating the logarithmic expression in the equation. The given equation is \[ -8 \log_{9}(x) = 16 \]To isolate \( \log_{9}(x) \), divide both sides of the equation by \(-8\) to get:\[ \log_{9}(x) = -2 \].

Step 2: Rewrite in Exponential Form

Use the definition of a logarithm to rewrite the equation in exponential form. The definition states that if \( \log_{b}(a) = c \), then \( a = b^c \). Applying this to \( \log_{9}(x) = -2 \), we have:\[ x = 9^{-2} \]

Step 3: Calculate the Exponential Expression

Calculate \( 9^{-2} \). Recall that a negative exponent means taking the reciprocal and squaring:\[ 9^{-2} = \frac{1}{9^2} = \frac{1}{81} \].

Step 4: Conclusion

Now that we have calculated the value of \( x \), substitute back to verify if necessary. The solution to the equation is \[ x = \frac{1}{81} \].

Key Concepts

Exponential Form

The concept of converting logarithmic equations into exponential form is a crucial step in solving many types of problems. Logarithms might seem intimidating at first, but breaking them down into simpler exponential forms can make the process straightforward. Here’s how it works: when you have a logarithmic equation like \(\log_{b}(a) = c\), you can convert it into an exponential statement which is \(a = b^c\). This transformation helps you see the relationship between the base \(b\), the exponent \(c\), and the result \(a\) with clarity.

• Think of the base of the logarithm (here, 9) as the base in the exponent expression.
• The log expression on the right of the equation, which is the exponent (\(c\)), tells you what power the base should be raised to.
• This results in the known expression \(x = 9^{-2}\) when applied to our example.

Applying this method involves recognizing the structure of the logarithm and confidently switching it into the language of exponents.

Isolating Logarithms

One pivotal step in solving logarithmic equations is isolating the logarithm itself, which often requires executing a sequence of algebraic manipulations. Isolation means getting the logarithmic part of the equation on one side all by itself. In the exercise, we start with the equation \(-8 \log_{9}(x) = 16\). To isolate the logarithmic expression \(\log_{9}(x)\), divide both sides of the equation by \(-8\):

• This simplification gives you \(\log_{9}(x) = -2\).
• Isolating makes it easier to apply logarithmic definitions and properties.
• Procedures like dividing or subtracting, depending on the equation, help remove coefficients that are hindering the isolation.

By isolating the logarithm, you effectively set the stage for the straightforward application of logarithmic principles, facilitating a smooth transition into rewriting the equation in its exponential form.

Negative Exponents

Understanding negative exponents is essential for interpreting and simplifying exponential equations, including those derived from logarithmic expressions. When faced with a negative exponent, it indicates that you need to take the reciprocal of the base. Taking the example \(9^{-2}\):

• The negative exponent here, \(-2\), means you take the reciprocal of 9, which turns \(9^2\) into \(\frac{1}{9^2}\).
• Calculating \(9^{2}\) gives us 81, making \(9^{-2} = \frac{1}{81}\).
• This reflects the general rule: \(b^{-n} = \frac{1}{b^n}\), which means you're inverting the base and then applying the positive exponent.

By mastering this concept, you can easily handle equations requiring you to manipulate expressions involving negative exponents, ensuring accurate and efficient solutions.