Week 6 - Graded Assignment - 6

Course: Jan 2026 - Mathematics I

Week 6 - Graded Assignment - 6

Last Submitted: You have last submitted on: 2026-03-25, 14:03 IST

Introduction

Question 1

• $\frac{\ln2}{\ln3-\ln2}$
• $\frac{\ln18}{\ln12-\ln8}$
• $\ln2$
• $\ln18$

Solution

Abstract Solution (Strategy)

1. [Base Normalization]: Divide the equation by the smallest power term to create a quadratic form.
2. [Decision rule]: Divide by $8^x$ and substitute $y = (3/2)^x$.

Procedure

• Step 1 – Divide: $\frac{18^x}{8^x} - \frac{12^x}{8^x} - 2 = 0 \implies (\frac{9}{4})^x - (\frac{3}{2})^x - 2 = 0$.
• Step 2 – Substitute: Let $y = (3/2)^x$. Then $y^2 - y - 2 = 0$.
• Step 3 – Factor: $(y-2)(y+1) = 0 \implies y = 2$ (since $y > 0$).
• Step 4 – Logarithm: $(3/2)^x = 2 \implies x \ln(1.5) = \ln 2 \implies x = \frac{\ln 2}{\ln 3 - \ln 2}$.
• Result: $\frac{\ln2}{\ln3-\ln2}$

Question 2

• $\log_5(1.5)$
• $0.25$
• $1.5$

Solution

Abstract Solution (Strategy)

1. [Arithmetic Mean]: If $AB = BC$, then $B$ is the arithmetic mean of $A$ and $C$.
2. [Formula]: $2B = A + C \implies 2\log_5(\text{term}_B) = \log_5(\text{term}_A) + \log_5(\text{term}_C)$.

Procedure

• Step 1 – Algebra: $2\log_5(3^x - 4.5) = \log_5(3) + \log_5(3^x - 2.25)$.
• Step 2 – Log Rules: $\log_5((3^x-4.5)^2) = \log_5(3(3^x-2.25))$.
• Step 3 – Quadratic: Let $y = 3^x$. $(y-4.5)^2 = 3y - 6.75 \implies y^2 - 12y + 27 = 0$.
• Step 4 – Roots: $y = 9$ or $y = 3$. If $y=3$, the term $3^x-4.5$ becomes negative (impossible for log). So $y=9$.
• Step 5 – Distance: $BC = \log_5(9-2.25) - \log_5(9-4.5) = \log_5(6.75) - \log_5(4.5) = \log_5(1.5)$.
• Result: $\log_5(1.5) \approx 0.2519$.

Question 3

Solution

Abstract Solution (Strategy)

1. [Define State]: $f(t) = 1000e^{-kt}$ is the number of people who have NOT heard the rumor.
2. [Decision rule]: Solve for $k$ using Day 1 data, then find $t$ for $f(t) = 500$.

Procedure

• Step 1 – Initial k: After 1 day, $40$ heard it, so $960$ haven't. $960 = 1000e^{-k} \implies e^{-k} = 0.96$.
• Step 2 – Target: We want half to have heard it, so $500$ haven't. $500 = 1000e^{-kt}$.
• Step 3 – Solve: $1/2 = (e^{-k})^t \implies 0.5 = (0.96)^t$.
• Step 4 – Log: $t = \frac{\ln(0.5)}{\ln(0.96)} \approx 16.98$.
• Result: 17 days.

Question 4

• $f$ is not a one to one function.
• $f$ is a one to one function.
• Range of $f$ is $\mathbb{R}$.
• $f$ is a bijective function.

Solution

Abstract Solution (Strategy)

1. [Injectivity Test]: Assume $f(a) = f(b)$ and check if $a = b$.
2. [Decision rule]: Cross-multiply and simplify.

Procedure

• Step 1 – Eq: $\frac{2e^a}{3e^a+1} = \frac{2e^b}{3e^b+1}$.
• Step 2 – X-mult: $2e^a(3e^b+1) = 2e^b(3e^a+1) \implies 6e^{a+b} + 2e^a = 6e^{a+b} + 2e^b$.
• Step 3 – Reduce: $e^a = e^b \implies a = b$.
• Step 4 – Surjectivity: As $x \to \infty$, $f(x) \to 2/3$. As $x \to -\infty$, $f(x) \to 0$. Range is $(0, 2/3)$, which is not $\mathbb{R}$.
• Result: One-to-one but not onto.

Question 5

• $\ln(\frac{2x}{2-3x})$
• $\ln(\frac{x}{2-3x})$
• $\ln(\frac{x}{2-x})$

Solution

Abstract Solution (Strategy)

1. [Isolate x]: Replace $f(x)$ with $y$ and solve for $x$.
2. [Decision rule]: Group $e^x$ terms after clearing the denominator.

Procedure

• Step 1 – Cross Multiply: $y = \frac{2e^x}{3e^x+1} \implies y(3e^x+1) = 2e^x$.
• Step 2 – Rearrange: $3ye^x + y = 2e^x \implies y = e^x(2-3y)$.
• Step 3 – Log: $e^x = \frac{y}{2-3y} \implies x = \ln(\frac{y}{2-3y})$.
• Step 4 – Swap: $f^{-1}(x) = \ln(\frac{x}{2-3x})$.
• Result: $\ln(\frac{x}{2-3x})$

Question 6

• $f(m) > f(n)$
• $f(m) < f(n)$
• $f(m) = f(n)$
• $f(m) \leq f(n)$

Solution

Abstract Solution (Strategy)

1. [Simplify Log]: Standardize the log bases.
2. [Decision rule]: Recognize the function as $u + 1/u$, which is strictly increasing for $u > 1$.

Procedure

• Step 1 – Simplify: $\log_{m+1}(0.01) = \log_{m+1}(10^{-2}) = \frac{-2}{\log_{10}(m+1)}$.
• Step 2 – Subst: $f(m) = \log_{10}(m+1) - \frac{1}{2}\left(\frac{-2}{\log_{10}(m+1)}\right) = \log_{10}(m+1) + \frac{1}{\log_{10}(m+1)}$.
• Step 3 – Behavior: Let $u = \log_{10}(m+1)$. Since $m > 9$, $u > \log_{10}(10) = 1$.
• Step 4 – Monotonicity: $u + 1/u$ is strictly increasing for $u > 1$ (derivative $1 - 1/u^2 > 0$).
• Result: $f(m) > f(n)$.

Question 7

• The jeweler sold 540 kg gold in 2019.
• The jeweler sold at least 730 kg gold in 2019.
• The jeweler sold at least 2 kg gold daily throughout the year 2019.
• The jeweler sold at least 10 kg gold daily throughout the year 2019.

Solution

Abstract Solution (Strategy)

1. [Extrema]: Use the AM-GM inequality on the simplified function.
2. [Decision rule]: Apply $u + 1/u \ge 2$.

Procedure

• Step 1 – Min Daily: $f(m) = u + 1/u$. Since $u > 0$ for all $m \ge 1$, we have $u + 1/u \ge 2$.
• Step 2 – Conclusion: Minimum daily gold sold is $2$ kg.
• Step 3 – Total Sales: If daily min is $2$ kg, then yearly total min $(365 \text{ days}) = 365 \times 2 = 730$ kg.
• Result: Min daily $2$ kg, Min total $730$ kg.

Question 8

• For logarithmic fall, $a=1.5$ and $h=2$.
• For exponential rise passing through $(10,0)$, $b=10$.
• Stock price in month 12 is ₹4,000.
• Without the vaccine, the investor would lose everything in month 12.

Solution

Abstract Solution (Strategy)

1. [Parameter Fit]: Use given points to find $a, b, h$.
2. [Decision rule]: Test the proposed values in the equations.

Procedure

• Step 1 – Rise: $y = 10^{x/b} - b$. Plug $(10, 0) \implies 0 = 10^{10/b} - b$. If $b=10$, $10^{10/10} - 10 = 0$. Confirmed.
• Step 2 – Fall: $y = -1.5 \log(x-2) + 1.5$. If $x=12$ (without vaccine), $y = -1.5 \log(10) + 1.5 = 0$. Confirmed.
• Step 3 – Month 12 (Rise): $y = 10^{12/10} - 10 = 10^{1.2} - 10 \approx 15.8 - 10 = 5.8$ (₹5,800). Not ₹4,000.
• Result: $a=1.5, h=2, b=10$ and loss at month 12 without vaccine are all correct.

Question 9

• Defined except in $(-2, -1) \cup (3, 4) \cup (5, 6)$.
• Invertible in $(-\infty, -2) \cup (1, 3) \cup (4, 5) \cup (6, \infty]$.
• Graph of a polynomial.
• Range could be $(-\infty, \infty)$.
• Graph could be $\log_{10}(1+(x+2)(x+1)(x-3)(x-4)(x-5)(x-6))$.
• Invertible in $[5, \infty)$.

Solution

Abstract Solution (Strategy)

1. [Log Zeros]: Intercepts of $P(x)$ become zeros of $\log(1+P(x))$.
2. [Decision rule]: Negative values of $P(x) + 1$ create domain gaps.

Procedure

• Step 1 – Roots: Intercepts are $-2, -1, 3, 4, 5, 6$.
• Step 2 – Logs: For $x$ where $P(x)+1 \le 0$, the function is undefined. This matches the visual gaps.
• Step 3 – Range: The vertical asymptotes at the edges of the blocks suggest the range goes from $-\infty$ to $\infty$.
• Result: Roots, gaps, and range all match the $\log_{10}(1+P(x))$ model.

Question 10

• $\log_5 2$ is rational.
• If $0 < b < 1$ and $0 < x < 1$, then $\log_b x > 0$.
• If $\log_3(\log_5 x) = 1$, then $x=625$.
• If $0 < b < 1$ and $0 < x < y$, then $\log_b x > \log_b y$.

Solution

Abstract Solution (Strategy)

1. [Log Base Inequalities]: Logarithms with bases between 0 and 1 are strictly decreasing functions.
2. [Decision rule]: Reverse the inequality when the base is a fraction ($0 < b < 1$).

Procedure

• Step 1 – Case A: If $x < 1$ and base $< 1$, then $\log_b x > \log_b 1 = 0$. (True)
• Step 2 – Case B: If $x < y$ and base $< 1$, the function is decreasing, so $\log_b x > \log_b y$. (True)
• Step 3 – Rationality: $\log_5 2 = p/q \implies 5^p = 2^q$ (impossible for integers $p,q > 0$). Irrational.
• Step 4 – Eq: $\log_3(\log_5 x) = 1 \implies \log_5 x = 3^1 = 3 \implies x = 5^3 = 125$.
• Result: Fractional bases reverse inequalities and flips the sign of logs for inputs $<1$.

Question 11

Solution

Abstract Solution (Strategy)

1. [Vertex Form]: For $ax^2+bx+c$, the max/min occurs at $x = -b/2a$.
2. [Decision rule]: Plug the vertex $x$ back into the function.

Procedure

• Step 1 – Vertex: $x = -4 / (2 \times -1) = 2$.
• Step 2 – Value: $g(2) = -(2)^2 + 4(2) + 77 = -4 + 8 + 77 = 81$.
• Result: 81

Question 12

Solution

Abstract Solution (Strategy)

1. [Monotonicity]: Logarithms are strictly increasing. Max output requires max input from $g(x)$.
2. [Decision rule]: Collapse the logs using the max of $g(x) = 81$.

Procedure

• Step 1 – Layer 1: $\log_3(81) = 4$.
• Step 2 – Layer 2: $\log_2(4) = 2$.
• Step 3 – Layer 3: $\log_2(2) = 1$.
• Result: 1

Question 13

Solution

Abstract Solution (Strategy)

1. [Condense]: Combine the logs on the left.
2. [Decision rule]: Drop the ln and solve the algebra.

Procedure

• Step 1 – Solve: $\ln(7(2-4x^2)) = \ln(14) \implies 14 - 28x^2 = 14$.
• Step 2 – Isolate: $-28x^2 = 0 \implies x = 0$.
• Step 3 – Check: $\ln(2 - 4(0)^2) = \ln 2$. Defined.
• Result: 1 solution ($x=0$).

Question 14

• Domain is $(-1, \infty)$.
• Domain is $(-\infty, 1)$.
• $f(x)$ is not one-to-one when $x \in (-1, 1)$.
• $f(x)$ is one-to-one when $x \in (-1, 1)$.

Solution

Abstract Solution (Strategy)

1. [Log Constraint]: Input $x+1$ must be $> 0$.
2. [Absolute Symmetry]: Abs functions flip the negative Y values, creating a "V" shape that fails injectivity.

Procedure

• Step 1 – Domain: $x+1 > 0 \implies x > -1$.
• Step 2 – One-to-one: $\log(x+1)$ hits 0 at $x=0$. For $x \in (-1, 0)$, $\log(x+1)$ is negative. Absolute value flips it to positive. Thus, for every positive value in the range, there are two $x$ values (one yielding $\log(x+1) = V$ and one yielding $\log(x+1) = -V$).
• Result: Domain $(-1, \infty)$ and NOT one-to-one.