Problem 26 - 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. $$1.2(0.9)^{x}=0.6$$

Step-by-Step Solution

Answer

x = 6.574 (to the nearest thousandth)

Step 1: Isolate the Exponential Term

Start by isolating the exponential term \((0.9)^x\) in the equation. Divide both sides by 1.2:\[(0.9)^x = \frac{0.6}{1.2}\]This simplifies to:\[(0.9)^x = 0.5\]

Step 2: Take the Logarithm of Both Sides

To solve for \(x\), take the logarithm of both sides of the equation. You can use either natural logarithms \(\ln\) or common logarithms \(\log\):\[x \cdot \log(0.9) = \log(0.5)\]

Step 3: Solve for x

Divide both sides by \(\log(0.9)\) to get the value of \(x\):\[x = \frac{\log(0.5)}{\log(0.9)}\]Use a calculator to compute the logarithms and find \(x\).

Step 4: Calculate Exact and Approximate Value of x

Using a calculator, compute the value of \(x\):\[ x = \frac{-0.3010}{-0.0458} \]This gives an approximate value of:\[x \approx 6.574\]

Step 5: Present the Solution

The exact form of the solution is \(x = \frac{\log(0.5)}{\log(0.9)}\) and the approximate solution, rounded to the nearest thousandth, is \(x \approx 6.574\).

Key Concepts

Logarithms

Logarithms are the mathematical tools we use to solve for the exponent in an equation where the variable is in the form of an exponent. When we need to isolate a variable that's in an exponent, we use logarithms to "bring down" the exponent. Let's consider the equation

• \((0.9)^x = 0.5\)

In this step, you can take the logarithm of both sides of the equation to make the exponent accessible. This is because of the property:

• \(\log(a^b) = b \cdot \log(a)\)

Using this property, your equation becomes

• \(x \cdot \log(0.9) = \log(0.5)\)

You could use either the common logarithm (base 10) or the natural logarithm (base \(e\)) for this purpose, as long as you apply the same type of logarithm consistently to both sides. Logarithms help us solve exponential equations by transforming them into a form where the variable can be easily manipulated.

Exact Form Solutions

An exact form solution is the form of the answer that's presented in fractional or symbolic form rather than a decimal approximation. It retains all information about the solution without rounding. In the problem we're looking at:

• \(x = \frac{\log(0.5)}{\log(0.9)}\)

This is the exact form of the solution because it precisely captures the relationship between the numbers, without any rounding. Exact solutions are important in mathematics as they provide the most accurate depiction of problems with mathematical beauty and elegance. This form is particularly important in theoretical studies and when further mathematical manipulation is required, ensuring no precision is lost to rounding errors.

Calculator Use

Calculators are essential tools in solving logarithmic and exponential equations, especially when working with non-terminating decimals. When you need to compute logarithms, particularly of a number that doesn't resolve cleanly, calculators can provide these values quickly and accurately. In the example, to find the approximate value of \(x\):

• First, find \(\log(0.5)\) which is approximately \(-0.3010\).
• Next, find \(\log(0.9)\) which is approximately \(-0.0458\).
• Then, divide these values: \(x = \frac{-0.3010}{-0.0458} \approx 6.574\).

Using a calculator for these steps, ensures that the division and the management of negative signs are both handled accurately. This allows you to expedite your evaluations and find solutions that are both exact and approximate, with minimal error.

Approximate Solutions

In contrast to exact form solutions, approximate solutions are numerical approximations expressed in decimal form, which are often rounded to a specified degree of accuracy. This involves using the previously computed logarithms to conclude with an answer like:

• \(x \approx 6.574\)

Obtaining an approximate solution involves calculating the exact solution using a calculator and rounding the result to the nearest thousandth, or as specified. It's useful in practical applications where a decimal answer is needed for real-world computations, such as finances or engineering, where approximate values can be substituted for precise calculations. This ensures clarity and functionality when exact numbers are not necessary or would make the process cumbersome. With practice, you'll become confident in determining when to use approximate solutions in real-world scenarios.