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

Step-by-Step Solution

Answer

Exact solution: \(x = \frac{3}{1 - 3\ln(2)}\), approximate solution: \(x \approx -2.780\).

Step 1: Rewrite the Equation in Logarithmic Form

Given the equation \[e^{x-3} = 2^{3x}\]we can take the natural logarithm (ln) of both sides to obtain:\[\ln(e^{x-3}) = \ln(2^{3x})\]Using the property of logarithms, \(\ln(a^b) = b \cdot \ln(a)\), we simplify to:\[(x-3) \cdot \ln(e) = 3x \cdot \ln(2)\]Since \(\ln(e) = 1\), this becomes:\[x - 3 = 3x \cdot \ln(2)\]

Step 2: Solve for x

Rearrange the equation \(x - 3 = 3x \cdot \ln(2)\) to solve for \(x\):\[x - 3x \cdot \ln(2) = 3\]Factor out \(x\) from the left side:\[x(1 - 3 \cdot \ln(2)) = 3\]Now, divide both sides by \((1 - 3 \cdot \ln(2))\) to isolate \(x\):\[x = \frac{3}{1 - 3 \cdot \ln(2)}\]

Step 3: Approximate the Solution

Use a calculator to approximate the solution. First, calculate \(\ln(2)\). Its approximate value is 0.693. Substitute this value into the equation:\[x = \frac{3}{1 - 3 \times 0.693}\]Calculate the denominator:\[1 - 3 \times 0.693 = 1 - 2.079 = -1.079\]Now calculate \(x\):\[x = \frac{3}{-1.079} \approx -2.780\]Hence, \(x \approx -2.780\) to the nearest thousandth.

Key Concepts

Logarithmic Form

Understanding the concept of logarithmic form is crucial when dealing with exponential equations, such as the one given in the exercise. The idea is to convert an exponential equation to a form that is easier to solve. In this scenario, we take both sides of the equation \[e^{x-3} = 2^{3x}\] and apply the natural logarithm, denoted as \(\ln\). This transforms our equation into a linear form involving logarithms: \[\ln(e^{x-3}) = \ln(2^{3x})\].
The properties of logarithms are very helpful here. The equation simplifies by applying the property that states \(\ln(a^b) = b \cdot \ln(a)\):

• The left side becomes \((x-3) \cdot \ln(e)\)
• The right side becomes \(3x \cdot \ln(2)\)

This property helps us linearize the exponential equation, making it more straightforward to isolate and solve for \(x\). This method is a systematic approach to transform the equation into a simpler form.

Exact Solution

An exact solution to an equation means expressing the solution algebraically in its simplest possible form without any approximation.
In the exponential equation discussed, after converting to logarithmic form, we isolate \(x\) from the equation: \[x - 3 = 3x \cdot \ln(2)\]. Next, we bring all terms containing \(x\) to one side, resulting in \[x - 3x \cdot \ln(2) = 3\]. By factoring out \(x\), we make the equation easier to handle: \[x(1 - 3 \cdot \ln(2)) = 3\]. Finally, by dividing both sides by \((1 - 3 \cdot \ln(2))\), we can solve for \(x\) exactly as: \[x = \frac{3}{1 - 3 \cdot \ln(2)}\]. Expressing the solution in this form allows us to manage it symbolically, giving us a precise answer before any numerical approximation.

Approximate Solution

Sometimes equations result in irrational numbers, which need to be approximated for practical purposes.
In our problem, we achieve an exact solution, \(x = \frac{3}{1 - 3 \cdot \ln(2)}\), but to find an approximate solution, we calculate the numerical value. We substitute the approximate value of \(\ln(2)\), which is about 0.693, into the equation.
After substituting, the expression becomes:

• \[1 - 3 \times 0.693 = -1.079\]
• Calculating \(x\): \[x = \frac{3}{-1.079} \]
• This simplifies to \(x \approx -2.780\)

This demonstrates the process of rounding off to the nearest thousandth, which is often necessary when dealing with real-world applications requiring numerical data.

Natural Logarithm

The natural logarithm, or \(\ln\), is a logarithm with base \(e\), where \(e\) is an irrational number approximately equal to 2.71828. This form of logarithm is particularly useful when dealing with exponential equations as \(e\) frequently appears in contexts involving growth and decay, such as compound interest and natural processes.
In our exercise, using \(\ln\) simplifies the initial equation \[e^{x-3} = 2^{3x}\] into a form where standard algebraic techniques can be applied.
The property that \(\ln(e^b) = b\) is key in decluttering and simplifying the left side of the equation. This enables us to handle equations naturally involving raises to the power of \(e\) with ease, converting exponential growth or decay contexts into manageable algebraic problems. The simplicity and practicality of the natural logarithm make it an indispensable tool in mathematics.