Problem 14 - 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. $$0.6^{x}=3$$

Step-by-Step Solution

Answer

Exact solution: \(x = \frac{\ln(3)}{\ln(0.6)}\); Approximate solution: \(x \approx -2.150\).

Step 1: Take the logarithm of both sides

To solve the equation \(0.6^x = 3\), we first take the natural logarithm (ln) of both sides to bring the exponent down: \[\ln(0.6^x) = \ln(3)\] Using the property of logarithms, we know \(\ln(a^b) = b \cdot \ln(a)\). This gives us: \[x \cdot \ln(0.6) = \ln(3)\]

Step 2: Solve for x

Now, we solve for \(x\) by isolating it on one side of the equation: \[x = \frac{\ln(3)}{\ln(0.6)}\] This expression gives us the exact form of the solution.

Step 3: Calculate the approximate solution

Using a calculator, compute the value of \(\ln(3)\) and \(\ln(0.6)\), then divide the two results: \[\ln(3) \approx 1.0986, \quad \ln(0.6) \approx -0.5108\] Plug these into the expression for \(x\): \[x \approx \frac{1.0986}{-0.5108} \approx -2.150\] Thus, \(x\) is approximately \(-2.150\) when rounded to the nearest thousandth.

Key Concepts

Logarithms

Logarithms are a key tool in solving exponential equations. They help us transform equations where the variable is an exponent into a more manageable algebraic form. To do this, we use logarithmic functions, such as the natural logarithm (often denoted as \(\ln\)).
The fundamental property of logarithms used here is \(\ln(a^b) = b \cdot \ln(a)\), which allows us to move the exponent down, making it easier to solve for the variable.
For example, in the equation \(0.6^x = 3\), we take the natural logarithm of both sides:

• \(\ln(0.6^x) = \ln(3)\)
• Through the property mentioned, it simplifies to \(x \cdot \ln(0.6) = \ln(3)\)

From here, we can isolate \(x\) by dividing both sides of the equation by \(\ln(0.6)\). This process converts the exponential equation into a simple division problem, making it much easier to solve.

Exact Solutions

An exact solution involves expressing the answer in its exact mathematical form, often without having to evaluate it into decimals unless specified.
When solving exponential equations, achieving an exact solution might mean resting the answer as a fraction involving logarithms.
For our given equation, after taking the natural logarithm of both sides and applying logarithmic properties, we arrive at the equation:

• \(x = \frac{\ln(3)}{\ln(0.6)}\)

This representation is the exact solution for the variable \(x\).
Exact solutions are essential because they allow us to understand and verify the precise relationships and values involved without rounding, which might introduce errors or inaccuracies.

Approximate Solutions

Approximate solutions are often necessary when dealing with irrational numbers or when a numeric answer is required.
Calculators help us compute these solutions to specified degrees of accuracy, like to the nearest thousandth.
To find an approximate solution, we first calculate the natural logarithms:

• \(\ln(3) \approx 1.0986\)
• \(\ln(0.6) \approx -0.5108\)

Substituting these values into our expression for \(x\), we get:

• \(x \approx \frac{1.0986}{-0.5108} \approx -2.150\)

Here, \(-2.150\) is the approximate value for \(x\).
Approximations are particularly useful in practical scenarios where exact values are less important than being in the ballpark in terms of size or magnitude.