Graph natural log of (e^x)/(e^x-1)

Problem

y=ln((ex)/(ex−1))

Solution

1. Identify the domain by ensuring the argument of the natural logarithm is positive. Since ex is always positive, we require ex−1>0 which implies ex>1 or x>0

2. Simplify the expression using the quotient rule for logarithms, ln(a/b)=ln(a)−ln(b)

y=ln(ex)−ln(ex−1)

3. Apply the inverse property ln(ex)=x to further simplify the function.

y=x−ln(ex−1)

4. Determine the vertical asymptote by finding where the argument of the logarithm approaches zero. As x→0 ex−1→0 so ln(ex−1)→−∞

(lim_x→0)(x−ln(ex−1))=∞

5. Determine the horizontal asymptote by examining the limit as x→∞ Rewrite the original expression as ln((ex)/(ex*(1−e(−x))))=ln(1/(1−e(−x)))

(lim_x→∞)(ln(1/(1−e(−x))))=ln(1)=0

6. Analyze the derivative to find the slope. Using y=x−ln(ex−1)

d(y)/d(x)=1−(ex)/(ex−1)

d(y)/d(x)=(ex−1−ex)/(ex−1)=(−1)/(ex−1)

7. Conclude the shape of the graph. Since x>0 ex−1>0 making the derivative always negative. The function is strictly decreasing from ∞ at x=0 toward the horizontal asymptote y=0 as x→∞

Final Answer

y=ln((ex)/(ex−1)), for *x>0

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