Problem 34 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and an approximation to four decimal places. See Example 4. $$ 6^{x}=7^{x-4} $$

Step-by-Step Solution

Answer

The exact solution is \(x = \frac{-4 \log(7)}{\log(6) - \log(7)}\) and approximately \(x \approx 49.5653\).

Step 1: Express the Equation with a Common Base

We start with the equation \(6^x = 7^{x-4}\). To solve it, we will take the logarithm of both sides. Using the property \(\log(a^b) = b\log(a)\), we get:\[x \log(6) = (x-4) \log(7)\]

Step 2: Distribute and Rearrange terms

Distribute the logarithm on the right side:\[x \log(6) = x \log(7) - 4 \log(7)\]Rearrange it to bring all terms involving \(x\) to one side:\[x \log(6) - x \log(7) = - 4 \log(7)\]

Step 3: Factor Out x

Factor \(x\) from the left side:\[x (\log(6) - \log(7)) = -4 \log(7)\]

Step 4: Solve for x

Solve for \(x\) by dividing both sides by \(\log(6) - \log(7)\):\[x = \frac{-4 \log(7)}{\log(6) - \log(7)}\]

Step 5: Calculate the Exact and Approximate Solution

Calculate \(\log(6)\) and \(\log(7)\), then substitute into the equation to find \(x\):\[\log(6) \approx 0.7782 \quad \text{and} \quad \log(7) \approx 0.8451\]Substitute these values:\[x = \frac{-4 \times 0.8451}{0.7782 - 0.8451} \approx 49.5653\]

Key Concepts

Logarithms: Unlocking the Power of Cooperation

Logarithms play a pivotal role in solving exponential equations, allowing us to work with exponents in a manageable way. When you see an equation where the variable is an exponent, like \(6^x = 7^{x-4}\), logs become our go-to tool. A logarithm answers the question: "To what power must a number (the base) be raised, to yield another number?" So, when we use a logarithm on both sides of an equation: \(\log(6^x) = \log(7^{x-4})\), it allows us to "bring down" the exponent, turning multiplication into addition or subtraction. This property is summarized as: \(\log(a^b) = b \log(a)\). When you take the log of both sides of the equation, you're capitalizing on this property, turning an exponential equation into a linear one, which is far easier to solve.

Exact Solution: Finding the Precise Answer

To find the exact solution of an exponential equation using logarithms, you should manipulate and simplify the equation as much as possible. Take the equation from the problem: \(6^x = 7^{x-4}\). By applying logarithms, it becomes \(x \log(6) = (x-4) \log(7)\). By distributing and re-arranging, we get all the terms with \(x\) on one side: \(x \log(6) - x \log(7) = -4 \log(7)\). The goal here is to isolate \(x\).

• First, factor \(x\) out of the equation to simplify it further: \(x(\log(6) - \log(7)) = -4 \log(7)\).
• This step makes it clear that \(x\) can be expressed as a fraction: \(x = \frac{-4 \log(7)}{\log(6) - \log(7)}\).

By using this expression, you have your exact solution in a mathematical form, ready to plug in any numbers.

Approximate Solution: Estimating with Precision

After finding the exact solution, often you’ll need to get an approximate numerical answer, especially if asked to round to a specific decimal place. In our case, consider the exact solution formula: \(x = \frac{-4 \log(7)}{\log(6) - \log(7)}\). The next step involves calculating the logarithms of the numbers involved.

• First, assess what \(\log(6)\) and \(\log(7)\) equal. For example, \(\log(6) \approx 0.7782\) and \(\log(7) \approx 0.8451\).
• Substitute these values into your exact solution equation to approximate \(x\): \(x = \frac{-4 \times 0.8451}{0.7782 - 0.8451}\).
• Solving this gives a numerical result, which in this case is approximately 49.5653 when rounded to four decimal places.

Approximations allow us to work practically with irrational numbers or expressions that are not easily simplified, providing useful numeric answers in real-world situations or when further calculations depend on a close estimate.