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

Problem

y=ln((e^x)/(e^x−1))

Solution

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

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

y=ln(e^x)−ln(e^x−1)

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

y=x−ln(e^x−1)

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

(lim_x→0^+)(x−ln(e^x−1))=∞

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

d(y)/d(x)=1−(e^x)/(e^x−1)

d(y)/d(x)=(e^x−1−e^x)/(e^x−1)=(−1)/(e^x−1)

7. Conclude the shape of the graph. Since x>0 e^x−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((e^x)/(e^x−1)), for *x>0

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