Problem 14 - A Graphical Approach to College Algebra

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. $$0.6^{x}=3$$

Step-by-Step Solution

Answer

Exact solution: \(x = \frac{\ln(3)}{\ln(0.6)}\); Approximate solution: \(x \approx -2.151\).

Step 1: Set up the equation

The given equation to solve is \(0.6^x = 3\). We need to find the value of \(x\) that satisfies this equation.

Step 2: Take the logarithm of both sides

Apply the logarithm to both sides of the equation to help isolate \(x\). Choose a logarithm base that is convenient, such as the natural logarithm (ln) or base 10 logarithm (log). Here, we use the natural logarithm: \[ \ln(0.6^x) = \ln(3) \]

Step 3: Apply the power rule for logarithms

The power rule of logarithms allows us to bring the exponent in front of the log expression. Thus, the equation becomes: \[ x \cdot \ln(0.6) = \ln(3) \]

Step 4: Solve for x

To solve for \(x\), divide both sides of the equation by \(\ln(0.6)\): \[ x = \frac{\ln(3)}{\ln(0.6)} \] This expression gives \(x\) in exact form.

Step 5: Use a calculator to find the approximate value of x

Calculate \(x\) using a calculator. Enter \(\ln(3)\) and \(\ln(0.6)\) to get: \[ x \approx \frac{1.0986}{-0.5108} \approx -2.151 \] Ensure the answer is rounded to the nearest thousandth.

Key Concepts

Logarithms

Logarithms are a powerful mathematical tool used to solve equations where the unknown variable is an exponent. When you have an equation like \(0.6^x = 3\), logarithms help us by transforming the multiplicative relationships into additive ones. This makes the equation easier to handle. The logarithm essentially answers the question: "To what power must the base be raised, to get a certain number?"
For example, if you take the log of both sides of \(0.6^x = 3\), the equation becomes \(\ln(0.6^x) = \ln(3)\). By using the properties of logarithms, particularly the power rule, we transform the equation to \(x \cdot \ln(0.6) = \ln(3)\). This simplifies the process of solving for the variable \(x\).
Logarithms can be in different bases: common logarithms (base 10) or natural logarithms (base \(e\)). Choosing the base depends on your calculator and the context of the problem.

Exact Form Solutions

An exact form solution retains all its components in a precise, mathematical format, without estimating or rounding. It's handy for communicating the exact nature of the answer and for further calculations where precision is crucial. For the equation \(0.6^x = 3\), solving gives us \(x = \frac{\ln(3)}{\ln(0.6)}\) as the exact form solution. This fraction is the solution expressed in terms of logarithms.
Exact solutions are beneficial because they remain accurate regardless of further manipulation or translation into other forms. It's what distinguishes this approach from approximation, which we will discuss next.

Approximation

Approximation is the method of finding a numerical estimate of an exact solution. After obtaining the exact form solution for \(x\), you may need to approximate it to understand its meaning in a practical sense or to make calculations simpler. In this example, we calculate the value of \(x\) as \(\approx -2.151\) after rounding to the nearest thousandth.
Approximations are useful in contexts where you're dealing with real-world measurements, requiring finite precision. Remember, the approximation should be clearly specified with its degree of accuracy, such as rounding to three decimal places in this case.

Calculator Use

Calculators are indispensable for evaluating logarithms, especially when working with irrational numbers or non-standard bases. When you need to convert an exact form solution to an approximation, a calculator can efficiently compute the logarithmic fractions.
Start by entering \(\ln(3)\), then \(\ln(0.6)\), and finally divide the two. Many scientific calculators allow you to directly input the fraction \(\frac{\ln(3)}{\ln(0.6)}\) to obtain \(x \approx -2.151\). Not only does this save time, but it also ensures accuracy in complex calculations.