Problem 32 - Algebra and Trigonometry

Question

Solve each exponential equation . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ 4 e^{7 x}=10,273 $$

Step-by-Step Solution

Answer

The solution to the equation \(4e^{7x} = 10273\) in decimal approximation, correct to two decimal places, is \(x \approx 2.15\).

Step 1: Isolate Exponential Term

Firstly, isolate the exponential term on one side of the equation. This can be done by dividing both sides of the equation \(4e^{7x} = 10273\) by 4. Doing so gives us \(e^{7x} = \frac{10273}{4}\)

Step 2: Apply Logarithm

To remove the exponent, apply the natural logarithm to both sides of the equation. Remember that one property of logarithms states that the natural log of e is 1: \(ln(e^{7x}) = ln(\frac{10273}{4})\). We can then leverage the property of logarithms that allows us to bring down the exponent, resulting in \(7x*ln(e) = ln(\frac{10273}{4})\). As mentioned, \(ln(e) = 1\), so our equation simplifies to \(7x = ln(\frac{10273}{4})\)

Step 3: Solve for x

Now we just solve for x by dividing both sides of the equation by 7: \(x = \frac{ln(\frac{10273}{4})}{7}\)

Step 4: Calculate Decimal Approximation

Finally, use a calculator to find the decimal approximation for \(x\), rounded to two decimal places: \(x \approx 2.15\)

Key Concepts

Natural Logarithms

Natural logarithms are based on the mathematical constant \( e \), which is approximately equal to 2.71828. The base of natural logarithms is \( e \), and they are symbolized by \( \ln \). One key property is that \( \ln(e) = 1 \). This property is extremely helpful when solving exponential equations as it simplifies the process of solving for variables in equations involving the exponential function \( e \). For example, when given an equation like \( e^{7x} = \frac{10273}{4} \), applying the natural logarithm on both sides allows us to break down the exponent. Natural logarithms help by transforming multiplicative relationships into additive ones.

• Great for simplifying complex equations
• Useful in calculus for integration and differentiation

In the present problem, using a natural logarithm allowed the exponent to be conveniently brought down, leading to a straightforward solution for \( x \).

Common Logarithms

Common logarithms use 10 as their base and are often represented without a base as \( \log \). Though not used in the specific solution for this exercise, common logarithms are another useful tool when solving exponential equations.They follow certain properties that make them helpful in various mathematical and real-world applications:

• \( \log_{10}(10) = 1 \)
• Transforms exponentiation into multiplication, facilitating complex calculations

Common logarithms shine in contexts involving base 10, like in sciences dealing with orders of magnitude. While natural logarithms were used in the current exercise, knowing about common logarithms broadens your toolkit for solving exponential problems. Common logs are alternately useful when dealing with decimal systems or when calculators default to base 10, offering easy calculations and familiar numerical results.

Decimal Approximation

Decimal approximation is the process of representing a number in a simpler form, usually to a set number of decimal places. In solving mathematical problems, exact values like \( \ln\left(\frac{10273}{4}\right) \) can be cumbersome to work with, making decimal approximation a valuable tool.Calculator precision allows you to simplify the result to any desired degree of accuracy. For example, in solving for \( x \), the calculation \( x = \frac{\ln\left(\frac{10273}{4}\right)}{7} \) results in the decimal \( x \approx 2.15 \) after approximation. Benefits of Decimal Approximation:

• Makes complex calculations more manageable
• Offers a quick way to grasp the size of a number
• Helps in comparing and understanding results easily

Always remember, while rounding or approximating, information might be lost, so it's essential to consider the context and how precise your approximation needs to be.