Problem 80 - College Algebra with Modeling and Visualization

Question

Solve each equation. Approximate answers to four decimal places when appropriate. (a) \(\log _{4} x=2\) (b) \(\log _{8} x=-1\) (c) \(\ln x=-2\)

Step-by-Step Solution

Answer

(a) 16, (b) 0.125, (c) 0.1353.

Step 1: Solve \( \log_{4} x = 2 \)

To solve \( \log_{4} x = 2 \), we need to convert the logarithmic equation into an exponential form. The equation \( \log_{4} x = 2 \) means that the base \( 4 \) raised to the power of \( 2 \) equals \( x \). Thus, write it as an exponential equation: \[ x = 4^2 \]Calculate \( 4^2 \) to find \( x \):\[ x = 16 \]

Step 2: Solve \( \log_{8} x = -1 \)

For \( \log_{8} x = -1 \), rewrite the equation in its exponential form. This means the base \( 8 \) raised to the power of \( -1 \) equals \( x \):\[ x = 8^{-1} \]Calculate \( 8^{-1} \) by taking the reciprocal of \( 8 \):\[ x = \frac{1}{8} \]Therefore, \( x = 0.125 \).

Step 3: Solve \( \ln x = -2 \)

The equation \( \ln x = -2 \) can be rewritten using the fact that \( \ln x \) is the natural logarithm, which has a base \( e \). Convert the logarithmic equation into its exponential form:\[ x = e^{-2} \]Calculate \( e^{-2} \):\[ x \approx 0.1353 \](Approximated to four decimal places.)

Key Concepts

Exponential Form

When tackling logarithmic equations, translating them into their exponential form can clarify the solution process. The exponential form is essentially the reverse operation of logarithms. For example, when you see an equation like \( \log_{b} x = y \), it can be interpreted in exponential form as \( x = b^y \). This means that the base \( b \) raised to the power \( y \) equals \( x \).
Using this method can simplify problems significantly, as it allows you to apply straightforward arithmetic to find solutions. For instance, if \( \log_{4} x = 2 \), converting to exponential form gives \( 4^2 = x \), so \( x = 16 \). Understanding this translation is essential, as it often serves as the first step in solving logarithmic equations.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special kind of logarithm that uses the base \( e \). The number \( e \) is an irrational constant approximately equal to 2.718. This particular type of logarithm is commonly used in calculus and certain exponential growth and decay problems. When you encounter an equation involving a natural logarithm, such as \( \ln x = -2 \), it can be converted into exponential form using the base \( e \).
For example, given \( \ln x = -2 \), the equivalent exponential form is \( x = e^{-2} \). This means that \( e \) raised to the power of \( -2 \) equals \( x \). Calculating \( e^{-2} \) will give you an approximate value of \( 0.1353 \), which, importantly, is accurate to four decimal places. Mastering natural logarithms is crucial for many mathematical and scientific applications due to their unique properties and frequent usage.

Exponential Equations

Exponential equations involve terms in which variables appear as exponents. Solving these equations often requires understanding both logarithmic and exponential concepts. When you encounter a situation like \( \log_{8} x = -1 \), rewrite it as an exponential equation: \( x = 8^{-1} \). This effectively flips the problem into a more manageable arithmetic operation.
Solving \( 8^{-1} \) involves calculating the reciprocal of 8, resulting in \( x = \frac{1}{8} \) or \( 0.125 \). This straightforward approach is facilitated by recognizing the inverse nature of logarithms and exponentials. Such translations from logarithmic to exponential form are invaluable tools in unraveling equations that appear complex at first glance. Remember, getting comfortable with these conversions can greatly enhance your problem-solving toolkit in mathematics.