Problem 70 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the exponential equation algebraically. Round your result to three decimal places. Use a graphing utility to verify your answer. $$250\left[\frac{(1+0.01)^{x}}{0.01}\right]=150,000$$

Step-by-Step Solution

Answer

The solution to this exponential equation up to three decimal places is \( x = 179.717 \).

Step 1: Simplify the exponential equation

Seeing that the equation has a fraction with an exponent, it's necessary to simplify it first. Divide both sides of the equation by 250 to clear the coefficient on the left side: \[ \frac{(1+0.01)^{x}}{0.01} = 600 \].

Step 2: Clear the denominator

Next, we clear the fraction on the left side by multiplying both sides of the equation by 0.01 and get the equation in the form \( (1.01)^x = 6 \).

Step 3: Solve for x using logarithms

To solve for x, take logarithms on both sides. You can use any base for the logarithms. Here, we use the natural logarithm (ln): \( ln((1.01)^x) = ln(6) \). Then, apply the logarithmic property \( ln(a^b) = b * ln(a) \) to bring down x: \( x * ln(1.01) = ln(6) \). Now, you can solve for x by isolating x on one side of the equation: \( x = \frac{ln(6)}{ln(1.01)} \).

Step 4: Calculate x and round to three decimal places

Using a calculator, find the natural logarithms and divide. Round off your solution to the third decimal place to get the final answer.

Step 5: Verification using a graphical method

Graph the function \( f(x) = (1.01)^x - 6 \) using a graphing utility. The x-intercept will confirm the solution found algebraically. Remember that the process of graphing depends on the specific utility being used.

Key Concepts

Solving Algebraically

When solving exponential equations algebraically, the goal is to isolate the exponent and use logarithms to solve for the variable. Let's break down the process using our example equation. Initially, we have:

• \( 250 \left[ \frac{(1 + 0.01)^{x}}{0.01} \right] = 150,000 \)

First, simplify by dividing both sides by 250, which gives us:

• \( \frac{(1 + 0.01)^{x}}{0.01} = 600 \)

Next, clear the fraction by multiplying both sides by 0.01:

• \( (1.01)^x = 6 \)

At this stage, the equation is in a simpler form, enabling us to proceed with taking logarithms. This process relies on your understanding of constants and coefficients. It's critical to perform operations step-by-step to ensure accuracy. Once simplified, we are ready to bring in logarithms for further steps.

Logarithms

Logarithms are a crucial tool for solving equations where the variable is an exponent. They allow us to "bring down" the exponent so that we can solve for the variable directly. Here's how it works in this case:
To solve \( (1.01)^x = 6 \), we take the natural logarithm on both sides:

• \( \ln((1.01)^x) = \ln(6) \)

We utilize the property of logarithms that allows us to move the exponent in front of the logarithm:

• \( x \cdot \ln(1.01) = \ln(6) \)

This conversion is key: rather than dealing with an exponent, we now have a solvable algebraic equation. To isolate \( x \), divide both sides by \( \ln(1.01) \):

• \( x = \frac{\ln(6)}{\ln(1.01)} \)

Finally, use a calculator to evaluate the natural logarithms and compute \( x \). Don't forget to round your result to three decimal places to meet the instructions given.

Graphing Utility Verification

Using a graphing utility extends the robustness of your solution by providing a visual confirmation. After finding your algebraic solution, graph the function:

• \( f(x) = (1.01)^x - 6 \)

The idea is to find the x-intercept of this function, which represents the solution where \( f(x) = 0 \). Follow these general steps:
1. Enter the function into your graphing utility.2. Adjust the viewing window to make sure the x-intercept is visible.3. Look for the point where the graph crosses the x-axis. This will be your solution for \( x \).
This method is incredibly useful as it helps confirm your algebraic calculations. It can also provide insights about the behavior of the function, but keep in mind that it is crucial to use an accurate scale to ensure the crossing point is visible and exact. Remember, different graphing utilities may require slightly different steps to perform this task.