
Question 43, 5.5.103
Part 1 of 4
HW Score: 43.11%, 19.4 of 45 points
 Points: 0 of 1

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Part 1
Show that ln left parenthesis 1 plus e Superscript 2 x Baseline right parenthesisequals2xplusln left parenthesis 1 plus e Superscript negative 2 x Baseline right parenthesis.
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Part 1
How can the argument of the logarithmic expression ln left parenthesis 1 plus e Superscript 2 x Baseline right parenthesis be rewritten to contain 1 plus e Superscript negative 2 x​? Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
Rewrite ln left parenthesis 1 plus e Superscript 2 x Baseline right parenthesis as the product of logarithmic functions 
  
enter your response here.
B.
Rewrite the argument of ln left parenthesis 1 plus e Superscript 2 x Baseline right parenthesis so that it is raised to the power 
  
enter your response here.
​(Type an integer or a​ fraction.)
C.
Factor out 
  
e Superscript negative 2 x from the argument of ln left parenthesis 1 plus e Superscript 2 x Baseline right parenthesis.


To show that \(\ln(1 + e^{2x}) = 2x + \ln(1 + e^{-2x})\), we need to manipulate the expression \(\ln(1 + e^{2x})\) so that it contains the term \(\ln(1 + e^{-2x})\).

Let's consider the options given:

### Option A:
Rewrite \(\ln(1 + e^{2x})\) as the product of logarithmic functions.

This option doesn't seem appropriate because the logarithm of a sum cannot be directly expressed as the product of logarithmic functions.

### Option B:
Rewrite the argument of \(\ln(1 + e^{2x})\) so that it is raised to the power.

This option is also not quite fitting for our needs because we need to manipulate the argument to get a form similar to \(\ln(1 + e^{-2x})\).

### Option C:
Factor out \(e^{-2x}\) from the argument of \(\ln(1 + e^{2x})\).

This seems like a promising approach. Let's see how this works in detail:

First, consider the original expression:
\[ \ln(1 + e^{2x}) \]

We want to express \(1 + e^{2x}\) in a way that includes \(1 + e^{-2x}\). Notice that if we multiply the numerator and denominator inside the logarithm by \(e^{-2x}\), we can achieve this:

\[
1 + e^{2x} = e^{2x}(e^{-2x} + 1)
\]

So, we rewrite the expression as:
\[
\ln(1 + e^{2x}) = \ln\left(e^{2x} \cdot (1 + e^{-2x})\right)
\]

Using the logarithm property \(\ln(ab) = \ln(a) + \ln(b)\), we get:
\[
\ln\left(e^{2x} \cdot (1 + e^{-2x})\right) = \ln(e^{2x}) + \ln(1 + e^{-2x})
\]

Since \(\ln(e^{2x}) = 2x\), we have:
\[
\ln(1 + e^{2x}) = 2x + \ln(1 + e^{-2x})
\]

Thus, the correct choice is:
### C. Factor out \(e^{-2x}\) from the argument of \(\ln(1 + e^{2x})\).

Would you like further details or have any questions about this solution?

Here are five related questions you might want to ask:
1. How do you simplify logarithmic expressions?
2. What are the properties of logarithms?
3. How can you use logarithms to solve exponential equations?
4. What is the change of base formula for logarithms?
5. How do you differentiate logarithmic functions?

**Tip:** Always remember to use the properties of logarithms to simplify expressions and solve equations effectively.

Understand how to manipulate logarithmic expressions with detailed steps. Learn how to prove ln(1 + e^(2x)) = 2x + ln(1 + e^(-2x)), involving logarithm properties and exponential functions.

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