Chapter 1: Numbers & Quantification

STEP 1 Apply the product law of exponents: $a^m · a^n = a^m+n$ with equal base $6561$:

$$6561^0.14 × 6561^0.11 = 6561^{0.14 + 0.11} = 6561^0.25 = 6561^1/4$$

STEP 2 Prime factorize 6561: $6561 = 3^8$ (since $3^4 = 81$ and $81^2 = 6561$):

$$(3^8)^1/4 = 3^{8 × (1/4)} = 3^2 = \mathbf{9}$$

EXPLANATION By fundamental definition of logarithm: $\log_b N = x \iff b^x = N$. For any positive real base $b = 5$, $5^x > 0$ for all real numbers $x \in \mathbb{R}$. No real power of 5 can equal zero. Therefore, $\log_5(0)$ is strictly undefined.

STEP 1 Express both sides with base 3: $(√3)^n = (3^1/2)^n = 3^n/2$.

STEP 2 Since $6561 = 3^8$, equate exponents: $n2 = 8 \implies n = 16$.

STEP 3 Evaluate $(n)^1/2 = √16 = \mathbf{4}$.

STEP 1 Substitute $2^3 = 8$: $5^x × 8 = 36 \implies 5^x = 368 = 4.5$.

STEP 2 Use index law $5^x+1 = 5^x × 5^1 = 4.5 × 5 = \mathbf{22.5}$.

STEP 1 Rewrite the LHS with outer power 40: $36¹²⁰ = 36^{3 × 40} = (36^3)⁴⁰$.

STEP 2 Since exponents are equal to 40, equate bases: $36^3 = 36 × x \implies x = 36^336 = 36^2 = \mathbf{1296}$.

STEP 1 Compute each radical power since $6561 = 3^8$:

• $(6561)^1/2 = (3^8)^1/2 = 3^4 = 81$
• $(6561)^1/4 = (3^8)^1/4 = 3^2 = 9$
• $(6561)^1/8 = (3^8)^1/8 = 3^1 = 3$

STEP 2 Sum all terms: $81 + 9 + 3 = \mathbf{93}$.

EXPLANATION Statement (c) claims $\log(3+4) = \log(3 × 4)$. However, $\log(3+4) = \log 7$, while $\log(3 × 4) = \log 12 \ne \log 7$. Logarithms do not distribute across addition ($\log(A+B) \ne \log A + \log B$).

STEP 1 Express RHS as powers of 5: $5 × 25 × 125 = 5^1 × 5^2 × 5^3 = 5¹⁺²⁺³ = 5^6$.

STEP 2 Equating powers: $a+b = 6$.

STEP 3 Evaluate $(a+b)^2 = 6^2 = \mathbf{36}$.

STEP 1 Factor out the common smaller power $7^-15$:

$$7^-14 - 7^-15 = 7^-15(7^{-14 - (-15)} - 1) = 7^-15(7^1 - 1) = 7^-15(6) = \mathbf{6 × 7^-15}$$

STEP 1 Factor out $3ⁿ⁺²$ from LHS: $3ⁿ⁺²(3^2 - 1) = 8$.

STEP 2 Simplify: $3ⁿ⁺²(9 - 1) = 8 \implies 3ⁿ⁺²(8) = 8 \implies 3ⁿ⁺² = 1 = 3^0$.

STEP 3 Equating exponents: $n + 2 = 0 \implies \mathbf{n = -2}$.

STEP 1 Express 1024 as a power of 2: $1024 = 2¹⁰$.

STEP 2 Evaluate the 10th root: $\sqrt[10]{1024} = (2¹⁰)^1/10 = 2^1 = 2$.

STEP 3 Therefore $2^x = 2^1 \implies x = 1$. Since $x = 1$ is absent from choices (b), (c), and (d), the answer is "None of these".

STEP 1 Invert negative powers: $3^-2 = 19$ and $5^-2 = 125$.

STEP 2 Subtract the fractions: $19 - 125 = 25 - 9225 = 16225$.

STEP 3 Take the square root: $(16225)^1/2 = √16√225 = \mathbf{415}$.

(i) 4: $4 = 2^2 \implies \mathbf{(100)_2}$

(ii) 49: $49 = 32 + 16 + 1 = 2^5 + 2^4 + 2^0 \implies \mathbf{(110001)_2}$

(iii) 267: $267 = 256 + 8 + 2 + 1 = 2^8 + 2^3 + 2^1 + 2^0 \implies \mathbf{(100001011)_2}$

(iv) 1024: $1024 = 2¹⁰ \implies \mathbf{(10000000000)_2}$

(v) 4103: $4103 = 4096 + 7 = 2¹² + (4 + 2 + 1) = 2¹² + 2^2 + 2^1 + 2^0 \implies \mathbf{(1000000000111)_2}$

(i) $(1000)_2$: $1 × 2^3 = \mathbf{8}$

(ii) $(11001)_2$: $2^4 + 2^3 + 2^0 = 16 + 8 + 1 = \mathbf{25}$

(iii) $(1110101)_2$: $2^6 + 2^5 + 2^4 + 2^2 + 2^0 = 64 + 32 + 16 + 4 + 1 = \mathbf{117}$

(iv) $(10011001)_2$: $2^7 + 2^4 + 2^3 + 2^0 = 128 + 16 + 8 + 1 = \mathbf{153}$

Apply definition $b^x = N \iff \log_b N = x$:

• (i) $10^3 = 1000 \implies \mathbf{\log₁₀ 1000 = 3}$
• (ii) $10^-2 = 0.01 \implies \mathbf{\log₁₀(0.01) = -2}$
• (iii) $(0.2)^3 = 0.008 \implies \mathbf{\log_0.2(0.008) = 3}$
• (iv) $4^-2 = 1/16 \implies \mathbf{\log_4(1/16) = -2}$
• (v) $(81)^3/4 = 27 \implies \mathbf{\log₈₁ 27 = 34}$

Apply equivalence $\log_b N = x \iff b^x = N$:

• (i) $\log_2 2 = 1 \implies \mathbf{2^1 = 2}$
• (ii) $\log_5(1/5) = -1 \implies \mathbf{5^-1 = 1/5}$
• (iii) $\log₁₀(0.001) = -3 \implies \mathbf{10^-3 = 0.001}$
• (iv) $\log₂₇ 9 = 2/3 \implies \mathbf{27^2/3 = 9}$
• (v) $\log_4 1 = 0 \implies \mathbf{4^0 = 1}$

(i) $\log_3 x = 2$: $x = 3^2 = \mathbf{9}$

(ii) $\log_0.2(x + 1) = -2$: $x + 1 = (0.2)^-2 = (15)^-2 = 5^2 = 25 \implies x = 25 - 1 = \mathbf{24}$

(iii) $\log_2(x^2 - 1) = 3$: $x^2 - 1 = 2^3 = 8 \implies x^2 = 9 \implies \mathbf{x = ± 3}$

(iv) $\log₁₀(x - 10) = 1$: $x - 10 = 10^1 = 10 \implies \mathbf{x = 20}$

Convert each number to scientific form $m × 10^p$ ($1 ≤ m < 10$). The characteristic is the power $p$:

• (i) $23.84 = 2.384 × 10^1 \implies \mathbf{1}$
• (ii) $384.76 = 3.8476 × 10^2 \implies \mathbf{2}$
• (iii) $9.857 = 9.857 × 10^0 \implies \mathbf{0}$
• (iv) $0.00035 = 3.5 × 10^-4 \implies \mathbf{-4 = \bar{4}}$
• (v) $0.00002356 = 2.356 × 10^-5 \implies \mathbf{-5 = \bar{5}}$

(i) 24.6: In log tables, Row 24 under Column 6 gives $\mathbf{0.3909}$.

(ii) 348.56: Round to 4 significant digits (348.6): Row 34, Col 8 is $0.5416$. Mean difference for 6 is $7$. Mantissa $= 0.5416 + 0.0007 = \mathbf{0.5423}$.

(iii) 2.768: Row 27, Col 6 gives $0.4409$. Mean difference for 8 is $13$. Mantissa $= 0.4409 + 0.0013 = \mathbf{0.4422}$.

(iv) 0.0056: Row 56, Col 0 gives $\mathbf{0.7482}$.

Combine Characteristic + Mantissa:

• (i) $\log 5 = 0 + 0.6990 = \mathbf{0.6990}$
• (ii) $\log 67 = 1 + 0.8261 = \mathbf{1.8261}$
• (iii) $\log 734.5 = 2 + 0.8660 = \mathbf{2.8660}$
• (iv) $\log(0.0056) = \bar{3} + 0.7482 = \mathbf{\bar{3}.7482 = -2.2518}$

(i) $Antilog(1.5647)$: Mantissa $0.5647 \to 3670$. Characteristic $= 1 \implies$ place decimal after $1+1=2$ digits: $\mathbf{36.70}$.

(ii) $Antilog(2.9031)$: Mantissa $0.9031 \to 8000$. Characteristic $= 2 \implies$ place decimal after 3 digits: $\mathbf{800.0}$.

(iii) $Antilog(2.4567)$: Mantissa $0.4567 \to 2862$. Characteristic $= 2 \implies$ place decimal after 3 digits: $\mathbf{286.2}$.

(iv) $\log x = -2.1304$: Convert to bar notation: $-2.1304 = -3 + (1 - 0.1304) = \bar{3}.8696$. Mantissa $0.8696 \to 7406$. Characteristic $= \bar{3} \implies$ two leading zeros after decimal: $\mathbf{0.007406}$.

STEP 1 Apply quotient law $x^ax^b = x^a-b$ inside each set of parentheses:

$$= (x^a-b)^a+b · (x^b-c)^b+c · (x^c-a)^c+a$$

STEP 2 Expand powers using difference of squares $(p-q)(p+q) = p^2 - q^2$:

$$= x^{a^2 - b^2} · x^{b^2 - c^2} · x^{c^2 - a^2}$$

STEP 3 Add exponents: $(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = 0$. Hence $x^0 = \mathbf{1}$.

STEP 1 Express numbers as powers of 7: $343 = 7^3$ and $16807 = 7^5$.

STEP 2 Equate exponents to form two linear equations:

• $x - y = 3 \quad \dots (1)$
• $x + y = 5 \quad \dots (2)$

STEP 3 Adding equations $(1)$ and $(2)$: $2x = 8 \implies \mathbf{x = 4}$ (and $y = 1$).

(a): Convert decimal to fraction: $0.04 = 4100 = 125 = 5^-2$.
$$(0.04)^-2.5 = (5^-2)^-5/2 = 5^{(-2) × (-5/2)} = 5^5 = \mathbf{3125}$$

(b): Convert all factors to base 6 ($36 = 6^2, 216 = 6^3$):
$$6^6.5 × (6^2)^4.5 \div (6^3)^4.5 = 6^6.5 × 6^9 \div 6^13.5 = 6^{6.5 + 9 - 13.5} = 6^2$$
Hence, the missing exponent is $\mathbf{2}$.

STEP 1 Factorize 196: $196 = 14^2$. Thus base $m = 14$ and power $n = 2$.

STEP 2 Substitute into $(m - 3)ⁿ⁺¹$:

$$(14 - 3)²⁺¹ = 11^3 = \mathbf{1331}$$

(a): Move coefficients into exponents: $2\log(5/9) = \log(25/81)$.

$$\log[7516 × 8125 × 32243] = \log[(7525) × (3216) × (81243)] = \log[3 × 2 × 13] = \mathbf{\log 2}$$

(b): $2\log(15/18) = \log(225/324)$. Combine terms:

$$\log[225324 × 16225 × 49] = \log[9 × 12 × 49] = \mathbf{\log 2}$$

STEP 1 Expand the quotient and product using logarithm rules:

$$\log(10xy^2) = \log 10 + \log x - \log(y^2) = 1 + \log x - 2\log y$$

STEP 2 Substitute $\log x = m+n$ and $\log y = m-n$:

$$= 1 + (m + n) - 2(m - n) = 1 + m + n - 2m + 2n = \mathbf{1 - m + 3n}$$

STEP 1 Apply the base power theorem $\log_b^k x = 1k\log_b x$ to convert all terms to base 2:

• $\log_4 x = \log_2^2 x = 12\log_2 x$
• $\log₁₆ x = \log_2^4 x = 14\log_2 x$

STEP 2 Factor out $\log_2 x$:

$$(1 + 12 + 14)\log_2 x = 214 \implies 74\log_2 x = 214$$

STEP 3 Multiply both sides by $47$: $\log_2 x = 3 \implies x = 2^3 = \mathbf{8}$.

(a): The exponent of the first term is $(p - q)(q + r - p) = pq + pr - p^2 - q^2 - qr + pq = 2pq + pr - qr - p^2 - q^2$. When summed cyclically across all three terms, every positive term cancels with an exact symmetric negative counterpart, yielding a total exponent of 0. Hence $x^0 = \mathbf{1}$.

(b): Multiply numerator and denominator of the 1st fraction by $x^p$, the 2nd by $x^q$, and the 3rd by $x^r$:

$$= x^px^p + x^q + x^r + x^qx^q + x^p + x^r + x^rx^r + x^p + x^q = x^p + x^q + x^rx^p + x^q + x^r = \mathbf{1}$$

STEP 1 Let $2^x = 3^y = 12^z = k$ (where $k > 0$).

Then $2 = k^1/x, \quad 3 = k^1/y, \quad 12 = k^1/z$.

STEP 2 Express the composite number 12 in prime factors: $12 = 2^2 × 3$.

Substitute their representations in terms of $k$:

$$k^1/z = (k^1/x)^2 · k^1/y = k^{2/x + 1/y}$$

STEP 3 Equate exponents on base $k$:

$$1z = 2x + 1y \implies 2x = 1z - 1y = y - zyz$$

STEP 4 Invert both sides: $x2 = yzy - z \implies \mathbf{x = 2yzy - z}$. (Hence Proved)

STEP 1 Express all terms in base 2 ($1024 = 2¹⁰, 16 = 2^4, 4 = 2^2$):

• Numerator: $(2¹⁰)^n × (2^2)²ⁿ⁺¹ = 2¹⁰ⁿ × 2⁴ⁿ⁺² = 2¹⁴ⁿ⁺²$
• Denominator: $(2^4)^n × (2^2)^n-1 = 2⁴ⁿ × 2^2n-2 = 2^6n-2$

STEP 2 Subtract denominator power: $2^{(14n+2) - (6n-2)} = 2⁸ⁿ⁺⁴ = (2^4)²ⁿ⁺¹ = \mathbf{16²ⁿ⁺¹}$ (or $4⁴ⁿ⁺²$).

STEP 1 By commutativity of multiplication, $ba = ab$.

STEP 2 Rewrite RHS: $(ba)^y-x = (ab)^-(x-y) = 1(ab)^x-y$.

STEP 3 Multiply both sides by $(ab)^x-y$: $[(ab)^x-y]^2 = 1 \implies (ab)^2(x-y) = (ab)^0$.

For $ab \ne 1$, $2(x - y) = 0 \implies \mathbf{x - y = 0}$ (i.e. $x = y$).

STEP 1 Express $x$ in base 10: $x = (10^2)^0.20 = 10^0.40$.

STEP 2 Compute $y^2$: $y^2 = (10^0.60)^2 = 10^1.20$.

STEP 3 Substitute into $x^z = y^2$: $(10^0.40)^z = 10^1.20 \implies 10^0.40z = 10^1.20$.

STEP 4 Equate powers: $0.40z = 1.20 \implies z = 1.200.40 = \mathbf{3}$.

(i): Let $N = (2.463)^3 × (72.42)^1/2(34.59)^5$.
$$\log N = 3\log(2.463) + 12\log(72.42) - 5\log(34.59)$$ $$= 3(0.3915) + 12(1.8598) - 5(1.5389) = 1.1745 + 0.9299 - 7.6945 = -5.5901 = \bar{6}.4099$$ $$N = Antilog(\bar{6}.4099) = \mathbf{2.570 × 10^-6}$$.

(ii): Let $K = √(238.2)^2 × 86.3492.3$.
$$\log K = 12[2\log(238.2) + \log(86.34) - \log(92.3)]$$ $$= 12[2(2.3770) + 1.9362 - 1.9652] = 12[4.7540 - 0.0290] = 12[4.7250] = 2.3625$$ $$K = Antilog(2.3625) = \mathbf{230.4}$$.

PART (i) Year 2029 implies $t = 2029 - 2010 = 19$ years.
$P = 4.7(1.02)¹⁹$. Taking base-10 logarithms:
$$\log P = \log 4.7 + 19\log(1.02) = 0.6721 + 19(0.0086) = 0.6721 + 0.1634 = 0.8355$$ $$P = Antilog(0.8355) ≈ 6.847 billion = \mathbf{6,847 million}.$$

PART (ii) At year 2020 ($t_1 = 10$), $P₂₀₂₀ = 4.7(1.02)¹⁰$. We seek year $t$ where $P_t = 2 × P₂₀₂₀$:
$$4.7(1.02)^t = 2 × 4.7(1.02)¹⁰ \implies (1.02)^t-10 = 2$$ Taking logarithms on both sides:
$$(t - 10)\log(1.02) = \log 2 \implies t - 10 = \log 2\log 1.02 = 0.301030.00860 ≈ 35 years$$ Hence $t = 10 + 35 = 45$ years after 2010, which corresponds to the calendar year $\mathbf{2055}$.

EXPLANATION Reason R is universally true for positive operands: $\log(MN) = \log M + \log N$ for $M, N > 0$. However, Assertion A is false because when $x < -3$, both $(x-3) < 0$ and $(x+3) < 0$. Even though their product $(x^2 - 9) > 0$ has a defined logarithm, the individual terms $\log(x-3)$ and $\log(x+3)$ are completely undefined in the real numbers! Equality only holds for $x > 3$.

EXPLANATION By product law: $a^{(p+q) + (p-q)} = a^p+q+p-q = a^2p$. Reason R provides the exact rule governing the multiplication of like bases. Both are true, and R is the correct explanation of A.

EXPLANATION $\log(xy) = 0 \implies xy = 10^0 = 1$. Reason R correctly describes the product rule and log identity that explains A.

EXPLANATION $13 = 8 + 4 + 1 = 1 · 2^3 + 1 · 2^2 + 0 · 2^1 + 1 · 2^0 = (1101)_2$. Reason R defines the fundamental principle of radix-2 binary decomposition.

EXPLANATION For $\log(\log x)$ to be defined, its inner argument $\log x$ must be strictly positive: $\log₁₀ x > 0 \implies x > 10^0 \implies x > 1$. Reason R gives the exact domain condition of the logarithmic function.