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Chapter 6: Problem 104

Solve each equation. $$ e^{-2 x+1}=13 $$

Short Answer

Expert verified
\( x = \frac{\text{ln}(13) - 1}{-2} \)

Step by step solution

01

Take the natural logarithm of both sides

Recall that to solve for the exponent when dealing with an equation in the form of Exponential functions, the natural logarithm (ln) can be applied to both sides of the equation. Applying the natural logarithm gives: \( \text{ln}(e^{-2x + 1}) = \text{ln}(13) \)
02

Simplify using the property of logarithms

Utilize the property of logarithms that \( \text{ln}(e^a) = a \). Applying this, we obtain: \( -2x + 1 = \text{ln}(13) \)
03

Isolate the term containing the variable

To isolate \( -2x \), subtract 1 from both sides: \( -2x = \text{ln}(13) - 1 \)
04

Solve for x

Divide both sides by -2 to solve for \( x \): \( x = \frac{\text{ln}(13) - 1}{-2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

natural logarithm
Understanding the natural logarithm is crucial for solving exponential equations such as the provided problem. The natural logarithm, denoted as ln, is the logarithm to the base of the mathematical constant e (approximately 2.71828). The natural logarithm of a number x is the power to which e must be raised to equal x. Remember that applying the natural logarithm can help us work with the exponent on the e, making the equation easier to solve. This is what we applied in Step 1, when we took the natural logarithm of both sides of the equation, resulting in:

\( \text{ln}(e^{-2x + 1}) = \text{ln}(13) \)
This transforms our exponential equation into a more manageable linear form.
logarithmic properties
Logarithmic properties are powerful tools to simplify and solve equations involving logarithms. One essential property is that the natural logarithm of an exponential function simplifies to the exponent itself:
  • \( \text{ln}(e^a) = a \)

Using this property, we simplified the equation in Step 2. Initially, we had the equation:
  • \( \text{ln}(e^{-2x + 1}) = \text{ln}(13) \)

Applying the property \( \text{ln}(e^a) = a \) transforms our equation into:

\( -2x + 1 = \text{ln}(13) \)
This substitution allows us to tackle the equation with simpler arithmetic operations instead of dealing with an exponent.
isolating variables
Isolating variables is a common technique in algebra used to solve equations. The goal is to get the variable we want to solve for alone on one side of the equation. This often involves several arithmetic operations to 'undo' whatever is affecting the variable.
In our problem, after simplifying the logarithm, we had:
  • \( -2x + 1 = \text{ln}(13) \)

The next step is to isolate the variable -2x. To do this, we subtract 1 from both sides:

\( -2x = \text{ln}(13) - 1 \)
Finally, we solve for x by dividing both sides by -2:

\( x = \frac{\text{ln}(13) - 1}{-2} \)
By carefully isolating x, we arrive at the solution to the equation. Understanding this method is key for solving algebraic equations efficiently.

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Most popular questions from this chapter

Write each expression as a single logarithm. \(\ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right)\)

Find the value of \(\log _{2} 4 \cdot \log _{4} 6 \cdot \log _{6} 8\).

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve: \(x-16 \sqrt{x}+48=0\)

Suppose that April has access to an investment that will pay 10 % interest compounded continuously. Which is better. to be given 1000 now so that she can take advantage of this investment opportunity or to be given $$ 1325$ after 3 years?

Solve each equation. $$ e^{2 x+5}=8 $$
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