Problem 12 - Intermediate Algebra

Question

For Problems \(1-14\), solve each exponential equation and express solutions to the nearest hundredth. $$ e^{x-1}=8.2 $$

Step-by-Step Solution

Answer

x ≈ 3.10.

Step 1: Take the Natural Logarithm

To solve the equation \(e^{x-1} = 8.2\), we take the natural logarithm of both sides to eliminate the exponential function. The equation becomes \(\ln(e^{x-1}) = \ln(8.2)\).

Step 2: Simplify the Left Side

The natural log of \(e^{x-1}\) simplifies to \(x - 1\) because \(\ln(e^y) = y\). So, the equation becomes \(x - 1 = \ln(8.2)\).

Step 3: Solve for x

To isolate \(x\), add \(1\) to both sides of the equation, resulting in \(x = \ln(8.2) + 1\).

Step 4: Calculate the Numerical Value

Use a calculator to find \(\ln(8.2)\) and then add \(1\) to this result. \(\ln(8.2) \approx 2.10413\). So, \(x \approx 2.10413 + 1 = 3.10413\).

Step 5: Round to the Nearest Hundredth

Round \(3.10413\) to the nearest hundredth. The hundredth place is the second digit after the decimal point. Rounding gives \(x \approx 3.10\).

Key Concepts

Natural Logarithm

The natural logarithm is a mathematical function often denoted as \(\ln(x)\). It is the inverse of the exponential function with base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. This means that if you take the natural logarithm of an exponential function \(e^y\), you simply get the exponent \(y\). For example, \(\ln(e^{x-1})\) simplifies to \(x-1\).
Natural logarithms are particularly useful when dealing with exponential equations or growth patterns, as they help simplify expressions by "undoing" the effect of exponentiation. This property of natural logarithms greatly helps in solving exponential equations, which we'll delve into more in the next section.
When solving exponential equations, applying a natural logarithm can be a first step to make the equation more manageable. You're essentially "flattening" the exponential curve to a straight line equation that you can solve using basic algebra.

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. In the context of exponential equations such as \(e^{x-1} = 8.2\), the goal is to isolate \(x\).
One effective method for solving this type of equation is by using the natural logarithm. By taking the natural logarithm of both sides, the equation \(e^{x-1} = 8.2\) is transformed into \(x-1 = \ln(8.2)\). This step simplifies the equation because \(\ln(e^{x-1}) = x-1\).

• Take the natural logarithm of both sides to deal with the exponent.
• Simplify the equation by setting the exponent equal to the logarithm of the other side.
• Isolate the variable by using algebraic manipulation, such as adding or subtracting numbers.

By following these steps, you can solve for \(x\) and find the solution to the equation.

Rounding Numbers

Rounding numbers is a process of adjusting numbers to make them simpler or to fit specific criteria, such as the nearest whole number, tenth, hundredth, etc. When rounding a decimal to the nearest hundredth, you focus on the second digit to the right of the decimal point.
For example, consider the number 3.10413. To round to the nearest hundredth, look at the hundredth place (\(0.10\)) and the next digit, which is \(4\). If this next digit is five or more, you round up the hundredth place by one. If it's less than five, you leave the hundredth place as is.

• Identify the digit in the hundredth place.
• Check the digit immediately after the hundredth place.
• If the next digit is 5 or greater, round up; otherwise, keep the digit unchanged.

Applying this rule to 3.10413 gives us 3.10 when rounded to the nearest hundredth, since the next digit is \(4\), which doesn't prompt a rounding up. Rounding is essential for getting a final result that is concise and easier to use, especially in practical applications.