Problem 19 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, rounded to four decimal places. \(e^{2 x+1}=200\)

Step-by-Step Solution

Answer

The solution is approximately \( x = 2.1517 \).

Step 1: Take Natural Logarithm

Apply the natural logarithm (ln) to both sides of the equation to eliminate the exponential function: \( \ln(e^{2x+1}) = \ln(200) \). This simplifies to \( 2x + 1 = \ln(200) \) because the natural logarithm and the exponential function are inverse functions.

Step 2: Isolate the Term with x

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

Step 3: Solve for x

Divide both sides by 2 to solve for \( x \): \( x = \frac{\ln(200) - 1}{2}\).

Step 4: Calculate and Round

Use a calculator to compute \( \ln(200) \), then substitute it into the expression \( x = \frac{\ln(200) - 1}{2} \) and round the result to four decimal places. This gives \( x \approx 2.1517 \).

Key Concepts

Natural Logarithm

Natural logarithm, often written as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). This concept is crucial in solving exponential equations because it helps transform an equation with an exponent into a more manageable form.

• The natural logarithm of \( e \) raised to any power simplifies the given power back to itself: \( \ln(e^x) = x \).
• This property is due to the fact that the natural logarithm is the inverse function of the exponential function \( e^x \).

Using the natural logarithm enables us to take complex exponential equations and convert them into linear ones, which are much simpler to solve. This transformation is a pivotal step in the process of dealing with equations involving exponential terms.

Inverse Functions

Inverse functions undo the action of a given function. When dealing with exponential equations, recognizing \( \ln \) and \( e^x \) as inverse functions is vital.

• An inverse function reverses the effect of its corresponding function. For instance, using \( \ln \) on \( e^x \) results in \( x \).
• This is why applying \( \ln \) to both sides of an exponential equation simplifies the problem considerably.

Applying this concept, in our problem, we go from \( e^{2x+1} = 200 \) to \( 2x + 1 = \ln(200) \) by taking \( \ln \) of both sides. This transformation is the key that unlocks the solution to the exponential equation.

Solving Exponential Equations

Solving exponential equations involves a few special techniques. First, it's important to recognize the interplay between different functions such as exponents and logarithms.

• Begin by applying the natural logarithm to both sides of the equation, as this helps to remove the exponential part.
• Next, isolate the variable by performing legal algebraic operations like subtracting or dividing by numbers to simplify.
• In this case, reducing \( 2x + 1 = \ln(200) \) to \( x = \frac{\ln(200) - 1}{2} \) shows how these algebraic operations function together to simplify the equation and make \( x \) the subject.

Each of these steps needs clear comprehension to systematically solve equations that initially appear complex due to their exponential nature.

Rounding Decimal Places

Rounding is essential when dealing with continuous quantities. Once you have calculated a value with many decimal places, rounding helps you provide a concise, understandable answer.

• Decide the number of places you need to round to. In many math problems, precision up to four decimal places suffices to give a clear representation of the value.
• Using a scientific calculator, compute the specific logarithmic value, then apply regular rounding rules (if the next digit is 5 or more, round up).
• For our solution, \( x \approx 2.1517 \), where the value of \( x \) has been accurately generated and then rounded to four decimal places.

Rounding ensures that the answer remains comprehensible and precise, balancing numerical accuracy with readability for practical application.