Problem 21 - 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. $$e^{x-3}=2^{3 x}$$

Step-by-Step Solution

Answer

Exact: \( x = \frac{3}{1 - 3\ln(2)} \); Approximate: \( x \approx -5.708 \).

Step 1: Take the Natural Logarithm on Both Sides

Start with the equation \(e^{x-3} = 2^{3x}\). To solve for \(x\), take the natural logarithm (ln) of both sides to eliminate the exponents: \(\ln(e^{x-3}) = \ln(2^{3x})\).

Step 2: Apply the Power Rule for Logarithms

By applying the power rule of logarithms \(\ln(a^b) = b \cdot \ln(a)\), we can rewrite both sides: \((x-3) \cdot \ln(e) = 3x \cdot \ln(2)\). Knowing that \(\ln(e) = 1\), we simplify to \(x-3 = 3x \cdot \ln(2)\).

Step 3: Solve for x

Rearrange the equation by moving terms involving \(x\) to one side: \(x - 3x \cdot \ln(2) = 3\). Factor out \(x\) from the left side: \(x(1 - 3\ln(2)) = 3\). Finally, solve for \(x\): \(x = \frac{3}{1 - 3\ln(2)}\).

Step 4: Exact Solution

The exact solution for \(x\) is given by the expression: \(x = \frac{3}{1 - 3\ln(2)}\).

Step 5: Approximate the Solution

Use a calculator to approximate \(\ln(2)\approx 0.693\). Substitute this into the expression for \(x\): \(x \approx \frac{3}{1 - 3 \times 0.693}\). Calculate the numerator and denominator to get \(x \approx -5.708\).

Step 6: Round to the Nearest Thousandth

Round the approximate solution to the nearest thousandth: \(x \approx -5.708\).

Key Concepts

Natural Logarithm

The natural logarithm is a fundamental concept in mathematics, especially when dealing with exponential equations like the one in our exercise. It is often denoted by the symbol \( \ln \). The base of the natural logarithm is the mathematical constant \( e \), which is approximately equal to 2.71828. This constant is special because it naturally arises in many growth/decay processes and calculus.

When solving exponential equations, taking the natural logarithm of both sides can help eliminate the exponential, allowing us to work with the exponents directly. For example, in the equation \( e^{x-3} = 2^{3x} \), applying \( \ln \) to both sides gives \( \ln(e^{x-3}) = \ln(2^{3x}) \). This step is crucial as it transforms the problem into one that is more manageable algebraically.

Power Rule for Logarithms

The power rule for logarithms is a handy tool when it comes to simplifying expressions where an exponent is being applied to a log argument. The rule states \( \ln(a^b) = b \cdot \ln(a) \). This simplification is important because it brings the exponent down in front of the logarithm, allowing us to handle it using basic algebra.

In our exercise, after taking natural logarithms, we use the power rule to get:

• \((x-3) \cdot \ln(e) = 3x \cdot \ln(2)\).

Since \( \ln(e) = 1 \), it simplifies further to \( x-3 = 3x \cdot \ln(2) \). This step makes it easier to isolate \( x \) and eventually find the solution. Understanding and using the power rule effectively can greatly simplify solving logarithmic and exponential problems.

Exact Solution

An exact solution refers to a solution derived purely through algebraic manipulation, without approximation. It's a precise expression that represents the solution in its most accurate form.

For the problem \( e^{x-3} = 2^{3x} \), once we used the natural logarithm and power rule, we rearranged the terms to solve for \( x \):

• \( x - 3x \cdot \ln(2) = 3 \)
• \( x(1 - 3\ln(2)) = 3 \)

To express \( x \) we isolate it on one side:

• \( x = \frac{3}{1 - 3\ln(2)} \)

This expression is your exact solution and shows \( x \) in terms of logarithms, not as a decimal or rounded number. It's important because it can offer more insight and flexibility for solving complex equations later.

Approximate Solution

An approximate solution is derived when exact forms are challenging to compute or interpret, typically by using numerical methods or calculators. Often, we need to convert logarithmic expressions into decimal forms for practical usage.

In our problem, while the exact solution for \( x \) is \( \frac{3}{1 - 3\ln(2)} \), it's often more useful to have a numerical value. Calculators help here as they can evaluate natural logarithms like \( \ln(2) \). With \( \ln(2) \approx 0.693 \), you calculate:

• \( x \approx \frac{3}{1 - 3 \times 0.693} \)

The approximate arithmetic gives \( x \approx -5.708 \), which is our decimal approximation. Rounding to the nearest thousandth gives accuracy needed in practical scenarios. Approximation is a key technique in many fields, particularly when precision isn't strictly necessary or feasible.