Problem 12 - A Graphical Approach to Precalculus with Limits

Question

Solve each exponential equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the nearest thousandth. Support your solutions by using a calculator. $$\left(\frac{1}{3}\right)^{x}=6$$

Step-by-Step Solution

Answer

Exact: \( x = \frac{\log_{10}(6)}{\log_{10}\left(\frac{1}{3}\right)} \); Approximate: \( x \approx -1.631 \).

Step 1: Write the Equation in Logarithmic Form

The given equation is \(\left(\frac{1}{3}\right)^{x}=6\). To solve for \(x\), rewrite the equation in logarithmic form. This gives: \[ x = \log_{\left(\frac{1}{3}\right)}(6) \]

Step 2: Apply Change of Base Formula

The change of base formula for logarithms allows any logarithm to be rewritten in terms of logs of any other base. Using the common logarithm (base 10), the formula is: \[ \log_{a}(b) = \frac{\log_{10}(b)}{\log_{10}(a)} \]Applying this to our equation: \[ x = \frac{\log_{10}(6)}{\log_{10}\left(\frac{1}{3}\right)} \]

Step 3: Evaluate Using a Calculator

Use a calculator to evaluate the expression:Compute \( \log_{10}(6) \) and \( \log_{10}\left(\frac{1}{3}\right) \):- \( \log_{10}(6) \approx 0.7781513 \)- \( \log_{10}\left(\frac{1}{3}\right) \approx -0.4771213 \)Plug these values into the formula:\[ x \approx \frac{0.7781513}{-0.4771213} \approx -1.631 \]

Step 4: Present the Solution

The exact solution is expressed in logarithmic form, while the approximation to the nearest thousandth is given as:- Exact form: \( x = \frac{\log_{10}(6)}{\log_{10}\left(\frac{1}{3}\right)} \)- Approximate form: \( x \approx -1.631 \)

Key Concepts

Logarithmic Form

When faced with exponential equations, we often need to find a more manageable form to solve for the unknown variable. This is where the concept of converting an equation into logarithmic form becomes essential. When we rewrite an exponential equation such as \(\left(\frac{1}{3}\right)^{x}=6\) into logarithmic form, it becomes \(x = \log_{\left(\frac{1}{3}\right)}(6)\).

This transformation process essentially asks, "What exponent do we put on \(\frac{1}{3}\) to get 6?" Using logarithms helps in isolating \(x\) by moving it out of the exponent. This change makes it increasingly easier to work with, especially when employing calculators or further algebraic manipulation.

Change of Base Formula

The change of base formula is a key tool when dealing with logarithms of bases that are not common or natural logarithms. This formula converts a logarithm with any base into a quotient of logarithms with a base we can compute, often base 10 or base \(e\). The formula is:

• \(\log_{a}(b) = \frac{\log_{10}(b)}{\log_{10}(a)}\) or \(\log_{a}(b) = \frac{\ln(b)}{\ln(a)}\).

Using this in our problem, we rewrite \(x = \log_{\left(\frac{1}{3}\right)}(6)\) by:

• \(x = \frac{\log_{10}(6)}{\log_{10}\left(\frac{1}{3}\right)}\).

This conversion allows for simple use of a scientific calculator, which can easily compute \(\log_{10}\) values. The adjusted formula makes tackling initially complex-looking logarithms straightforward.

Solving Equations

Once the logarithmic form and change of base formula are applied, solving the equation becomes a task of evaluating expressions. Our transformed equation \(x = \frac{\log_{10}(6)}{\log_{10}\left(\frac{1}{3}\right)}\) leads us to plug these terms into a calculator. By computing each logarithm:

• \(\log_{10}(6) \approx 0.7781513\)
• \(\log_{10}\left(\frac{1}{3}\right) \approx -0.4771213\)

and then performing the division:

• \(x \approx \frac{0.7781513}{-0.4771213} \approx -1.631\)

This step consolidates our understanding of how changing the complex original equation into a more manageable format allows for efficient solutions.

Exact Form and Approximation

After solving exponential equations, presenting the solution in exact form and approximate form is vital for comprehensive understanding. The exact form retains the logarithmic expression, offering insights into the relationship between values:

• Exact form: \(x = \frac{\log_{10}(6)}{\log_{10}\left(\frac{1}{3}\right)}\)

This form is crucial in mathematical proofs or when precision is necessary. However, as mathematicians, sometimes we need a practical number. Thus, we approximate the value to a more digestible format:

• Approximate form: \(x \approx -1.631\)

Expressing solutions in both forms ensures thorough understanding—exact values reveal precise calculations, while approximations provide easy-to-interpret numeric answers.