📝 Exponential and logarithmic growth applications (13 MCQs)

📖 From Calculus • 1. Basics before calculus • 13 questions available

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What is Exponential and logarithmic growth applications?

Definition:
Exponential growth ( $y = a e^{kt}$ ) models rapid increase, while logarithmic growth ( $y = a \ln(t) + b$ ) models slow increase, with applications in populations, finance, algorithms, and sensory perception, respectively.

Example:
A bacteria culture doubles every hour: $N(t) = N_0 2^t$ ; learning curve: $y = 5 \ln(t) + 10$ shows diminishing returns in skill improvement over time.

Reason:
Understanding these growth types helps in predicting trends, optimizing resources, and interpreting data in biology, economics, and computer science (e.g., time complexity).

Easy

Medium

Hard

📝 All Exponential and logarithmic growth applications MCQs

Q1. If $y = e^{x}$ exceeds 1000, which inequality must hold for $x$ ?

x > \ln 1000

✅

x < \ln 1000

x > 1000

x < 1000

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since $e^{x}$ is strictly increasing, the inequality $e^{x}>1000$ is equivalent to $x>\ln 1000$ . Any smaller $x$ would give a value of $e^{x}$ less than 1000, so the only valid statement is $x>\ln 1000$ .

Q2. Given that $\ln x > 5$ , which statement is always true?

x > 5

x > e^{5}

✅

x < e^{5}

x < 5

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The natural logarithm function is monotonic increasing, so $\ln x > 5$ implies $x$ must be greater than the number whose logarithm is 5, namely $e^{5}$ . Any value less than $e^{5}$ would produce a logarithm at most 5, contradicting the premise.

Q3. Suppose $f(x)=e^{x}$ and $g(x)=\ln x$ . If for some $a>0$ we have $f(a)=g(a^{2})$ , which must be true about $a$ ?

a>1

a=1

0<a<1

D.No solution ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Setting $e^{a}=\ln(a^{2})$ leads to $e^{a}=2\ln a$ . For $a>0$ , the left side grows exponentially while the right side grows only logarithmically, and at $a=1$ the equality fails. Consequently the equation has no real solution, making “No solution” the correct choice.

Q4. If the function $h(x)=\ln(e^{x}+1)$ is increasing for all real $x$ , what can be inferred about its derivative $h'(x)$ ?

h&#039;(x)=e^{x}+1

h&#039;(x)=\frac{1}{e^{x}+1}

h&#039;(x)=\frac{e^{x}}{e^{x}+1}>0

✅

h&#039;(x)=\ln(e^{x}+1)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Differentiating gives $h'(x)=\frac{e^{x}}{e^{x}+1}$ . Because both numerator and denominator are positive for all real $x$ , the fraction is positive, confirming that $h(x)$ is indeed increasing everywhere. The other options either misrepresent the derivative or give a non‑positive expression.

Q5. Which function grows faster as $x\to\infty$ : $e^{x}$ or $x^{5}$ ?

x^{5}

grows faster

e^{x}

grows faster ✅

C.Both grow at the same rate

D.Cannot be compared

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Exponential growth outpaces any polynomial. As $x$ becomes large, the ratio $\frac{e^{x}}{x^{5}}$ tends to infinity, showing that $e^{x}$ dominates $x^{5}$ . Hence $e^{x}$ grows faster, which is captured by option B.

Q6. Evaluate $\displaystyle\lim_{x\to\infty}\frac{\ln x}{e^{x}}$ .

\infty

B.1

C.Does not exist

D.0 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The exponential function grows much more rapidly than the logarithm. Applying L'Hôpital's Rule once gives $\lim_{x\to\infty}\frac{1/x}{e^{x}}=0$ because the numerator approaches zero while the denominator grows without bound. Therefore the original limit equals 0.

Q7. For which value of $k$ does the function $f(x)=e^{x}-k\ln x$ have exactly one critical point on $(0,\infty)$ ?

A.k = 0

B.k < 0

C.Any positive real number ✅

D.No such k

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Setting the derivative to zero yields $e^{x}=k/x$ or $k=x e^{x}$ . The function $x e^{x}$ is strictly increasing from 0 to $\infty$ , so for every positive $k$ there is exactly one solution, giving a single critical point. Hence any positive $k$ works.

Q8. Consider $p(x)=e^{2x}$ and $q(x)=\ln(x^{2})$ . Which statement correctly describes their relative growth for large $x$ ?

q(x)

eventually dominates

p(x)

B.Neither dominates the other

C.They grow at comparable rates

p(x)

eventually dominates

q(x)

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Because $e^{2x}$ grows exponentially while $\ln(x^{2})=2\ln x$ grows only logarithmically, the exponential term outpaces the logarithmic term for sufficiently large $x$ . Thus $p(x)$ dominates $q(x)$ as $x\to\infty$ .

Q9. Solve for $x$ : $e^{2x}=7$ .

x = \ln 7

x = \frac{1}{2}\ln 7

✅

x = \frac{1}{2}\ln 14

x = \ln \sqrt{7}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Taking natural logs of both sides gives $2x=\ln 7$ . Dividing by 2 yields $x=\frac{1}{2}\ln 7$ . The other options either miss the factor of 2 or misapply logarithmic identities.

Q10. If $y=\ln(x)$ is shifted 3 units up, what is the new equation?

y = \ln(x+3)

y = \ln(x-3)+3

y = e^{x}+3

y = \ln(x)+3

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: A vertical shift adds a constant to the entire function. Raising the graph by 3 units changes the equation to $y=\ln(x)+3$ . The other choices represent horizontal shifts or unrelated transformations, which do not match a pure upward translation.

Q11. A population follows $P(t)=P_0 e^{kt}$ . If it doubles in 5 years, what is the expression for $k$ using natural logarithms?

k = 5\ln 2

k = \frac{2}{\ln 5}

k = \frac{\ln 2}{5}

✅

k = \ln(2/5)

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Doubling means $P(t)=2P_0$ . Substituting gives $2P_0=P_0 e^{k\cdot5}$ → $e^{5k}=2$ . Taking natural logs: $5k=\ln 2$ → $k=\frac{\ln 2}{5}$ . This isolates $k$ in terms of $\ln$ .

Q12. Which expression represents the inverse of $f(x)=e^{x}$ ?

f^{-1}(x)=e^{x}

f^{-1}(x)=\ln x

✅

f^{-1}(x)=\frac{1}{e^{x}}

f^{-1}(x)=\log_{10} x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The inverse function reverses the effect of $e^{x}$ . Solving $y=e^{x}$ for $x$ yields $x=\ln y$ . Hence the inverse is $f^{-1}(x)=\ln x$ . The other options either repeat the original function or use unrelated operations.

Q13. What is the domain of the natural logarithm function $\ln x$ ?

x\ge 0

B.All real numbers

x<0

x>0

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The natural logarithm is defined only for positive arguments because $\ln x$ represents the exponent to which $e$ must be raised to obtain $x$ . Therefore the domain is the set of all real numbers greater than zero, i.e., $x>0$ .

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