Problem 31 - 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. $$5(1.015)^{x-1980}=8$$

Step-by-Step Solution

Answer

The solution is \(x \approx 2011.500\).

Step 1: Isolate the Exponential Expression

To solve the equation \(5(1.015)^{x-1980} = 8\), the first step is to isolate the exponential expression. To do this, divide both sides of the equation by 5: \[(1.015)^{x-1980} = \frac{8}{5}\] Simplifying gives: \[(1.015)^{x-1980} = 1.6\]

Step 2: Use Logarithms to Solve for x

Take the natural logarithm (ln) of both sides of the equation to solve for \(x-1980\): \[\ln((1.015)^{x-1980}) = \ln(1.6)\] Using the property of logarithms \(\ln(a^b) = b\ln(a)\), the equation becomes: \[(x-1980) \cdot \ln(1.015) = \ln(1.6)\] Now solve for \(x-1980\) by dividing both sides by \(\ln(1.015)\): \[x-1980 = \frac{\ln(1.6)}{\ln(1.015)}\]

Step 3: Solve for x

Calculate the value of \(\frac{\ln(1.6)}{\ln(1.015)}\) using a calculator. First, find \(\ln(1.6)\) which is approximately 0.470. Then, find \(\ln(1.015)\) which is approximately 0.01492. Now compute: \[x-1980 = \frac{0.470}{0.01492} \approx 31.50\] Add 1980 to both sides to find \(x\): \[x = 31.50 + 1980 = 2011.50\]

Step 4: Express the Solution

The exact form of the solution in terms of \(x\) is: \[x = 1980 + \frac{\ln(1.6)}{\ln(1.015)}\] The approximate solution to the nearest thousandth is: \[x \approx 2011.500\]

Key Concepts

Logarithms

Logarithms can be a powerful tool in solving exponential equations. When an equation involves exponents, taking the logarithm of both sides helps us bring the exponent down where it can be dealt with algebraically. In our equation, after isolating the exponential part, we used the natural logarithm (ln) to simplify our calculations.

• This is based on the property that \(\ln(a^b) = b\ln(a)\).
• Using logarithms allowed us to transform the exponential equation into a linear one.

This makes it easier to solve for the unknown variable, which in this case was \(x - 1980\). If we didn't use logarithms, finding the exact value of \(x\) would be much more challenging.

Exact Solutions

When solving exponential equations, finding exact solutions means representing the solution in its most precise form.

• An exact solution is expressed in terms of integers, fractions, or irrational numbers.
• For our problem, the exact form of the solution is expressed as: \[x = 1980 + \frac{\ln(1.6)}{\ln(1.015)}\]

This form isn't a round number but still represents a precise mathematical solution. It retains all the necessary detail, unlike approximate solutions that round numbers off.

Approximation

Often in mathematics, we need to provide numbers in a more digestible format. This is where approximation comes in. Approximations simplify complex results so that they are easier to read and understand, though they are not exact.

• For our solution, we needed to approximate to the nearest thousandth.
• In practice, our solution became:\[x \approx 2011.500\]

This indicates that while the original calculation may have had many decimal places, we shortened it to just three decimal places after the point, making it more user-friendly.

Calculator Usage

Using a calculator is essential when solving equations like the one in this exercise. It helps to compute the logarithms and other complex operations efficiently and accurately.

• To compute \(\ln(1.6)\) and \(\ln(1.015)\), a calculator is indispensable.
• A calculator also lets us handle division and addition very quickly.
• Without a calculator, we might spend unnecessary time doing manual calculations and increasing the chance of error.

In exams or real-life scenarios, knowing how to use your calculator effectively can save lots of time and also improve your precision in finding solutions.