Problem 17 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, correct to four decimal places. $$ e^{2 x+1}=200 $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 2.1492\).

Step 1: Take Natural Logarithm on Both Sides

To solve the exponential equation \(e^{2x+1} = 200\), start by taking the natural logarithm (\(\ln\)) on both sides of the equation to remove the exponential function. The equation becomes:\[\ln(e^{2x+1}) = \ln(200)\]

Step 2: Apply Logarithm Rules

Apply the rule of logarithms, \(\ln(a^b) = b\ln(a)\), to simplify the left side. Since \(e\) is the base of the natural logarithm (\(\ln(e) = 1\)), we have:\[2x + 1 = \ln(200)\]

Step 3: Isolate the Variable Term

Subtract 1 from both sides of the equation to isolate the term with \(x\):\[2x = \ln(200) - 1\]

Step 4: Solve for x

Divide both sides of the equation by 2 to solve for \(x\):\[x = \frac{\ln(200) - 1}{2}\]

Step 5: Calculate the Value

Use a calculator to find \(\ln(200)\) and substitute back to find \(x\). First compute \(\ln(200) \approx 5.2983\), then:\[x = \frac{5.2983 - 1}{2} \approx \frac{4.2983}{2}\]\[x \approx 2.1492\]

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \(\ln\), is a special type of logarithm where the base is the mathematical constant \(e\), approximately equal to 2.71828. It is commonly used in solving exponential equations because it conveniently simplifies expressions involving \(e\). When we take the natural logarithm of an exponential expression like \(e^{2x+1}\), it cancels out the \(e\), making the equation linear and easier to solve. In our problem, applying the natural logarithm on both sides gives us:\[\ln(e^{2x+1}) = \ln(200)\]Since \(\ln(e) = 1\), we simplify the left-hand side to obtain:\[2x+1 = \ln(200)\]This step is crucial as it transforms the equation from an exponential to a simple linear form.

Logarithm Rules

Logarithm rules are essential tools for manipulating expressions involving logarithms. A key rule applied in our problem is \(\ln(a^b) = b \cdot \ln(a)\), which allows us to bring down exponents as coefficients. This rule makes it straightforward to simplify terms in exponential equations.

• For the equation \(e^{2x+1}\), \(2x+1\) comes down as a factor by the rule.
• We then have \(\ln(e^{2x+1}) = 2x + 1\) because \(\ln(e) = 1\).

Understanding these rules facilitates solving equations where variables are in the exponent, reducing them to simpler algebraic operations.

Solving for Variables

Once the logarithmic transformation is complete, we need to solve for the variable \(x\). In our example, the equation is simplified to:\[2x + 1 = \ln(200)\]To isolate \(x\), follow these steps:

• Subtract 1 from both sides: \(2x = \ln(200) - 1\).
• Divide the entire equation by 2: \(x = \frac{\ln(200) - 1}{2}\).

This process eliminates constants and isolates the variable on one side of the equation, allowing us to directly solve for \(x\). This step is a fundamental part of algebraic manipulation and variable isolation.

Calculator Usage

Accurate calculation is crucial for solving exponential equations, especially when it involves logarithms. To get the value of \(x\) in our problem, you'll need a calculator capable of computing logarithms.Here’s how you use the calculator:

• Input \(\ln(200)\) into the calculator to get approximately 5.2983.
• Subtract 1 from this value: \(5.2983 - 1 = 4.2983\).
• Divide by 2 to get the final \(x\) value: \(\frac{4.2983}{2} \approx 2.1492\).

Ensure your calculator is in the correct mode for logarithmic calculations, and remember, the use of the natural logarithm \(\ln\) is key here, not the base-10 logarithm. This careful usage results in accurate and precise solutions.