Chapter 1: Numbers & Quantification

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

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

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

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

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: $(\sqrt{3})^n = (3^{1/2})^n = 3^{n/2}$.

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

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

STEP 1 Substitute $2^3 = 8$: $5^x \times 8 = 36 \implies 5^x = \frac{36}{8} = 4.5$.

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

STEP 1 Rewrite the LHS with outer power 40: $36^{120} = 36^{3 \times 40} = (36^3)^{40}$.

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

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

STEP 1 Express RHS as powers of 5: $5 \times 25 \times 125 = 5^1 \times 5^2 \times 5^3 = 5^{1+2+3} = 5^6$.

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 $7^{-15}$:

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

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

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

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

STEP 1 Express 1024 as a power of 2: $1024 = 2^{10}$.

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

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: $3^{-2} = \frac{1}{9}$ and $5^{-2} = \frac{1}{25}$.

STEP 2 Subtract the fractions: $\frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225}$.

STEP 3 Take the square root: $\left(\frac{16}{225}\right)^{1/2} = \frac{\sqrt{16}}{\sqrt{225}} = \mathbf{\frac{4}{15}}$.

(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: $1024 = 2^{10} \implies \mathbf{(10000000000)_2}$

(v) 4103: $4103 = 4096 + 7 = 2^{12} + (4 + 2 + 1) = 2^{12} + 2^2 + 2^1 + 2^0 \implies \mathbf{(1000000000111)_2}$

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

(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) $10^3 = 1000 \implies \mathbf{\log_{10} 1000 = 3}$
• (ii) $10^{-2} = 0.01 \implies \mathbf{\log_{10}(0.01) = -2}$
• (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) $(81)^{3/4} = 27 \implies \mathbf{\log_{81} 27 = \frac{3}{4}}$

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) $\log_{10}(0.001) = -3 \implies \mathbf{10^{-3} = 0.001}$
• (iv) $\log_{27} 9 = 2/3 \implies \mathbf{27^{2/3} = 9}$
• (v) $\log_4 1 = 0 \implies \mathbf{4^0 = 1}$

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

(ii) $\log_{0.2}(x + 1) = -2$: $x + 1 = (0.2)^{-2} = \left(\frac{1}{5}\right)^{-2} = 5^2 = 25 \implies x = 25 - 1 = \mathbf{24}$

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

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

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

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

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

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

(iii) $\text{Antilog}(2.4567)$: 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 $\frac{x^a}{x^b} = x^{a-b}$ inside each set of parentheses:

$$= (x^{a-b})^{a+b} \cdot (x^{b-c})^{b+c} \cdot (x^{c-a})^{c+a}$$

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

$$= x^{a^2 - b^2} \cdot x^{b^2 - c^2} \cdot x^{c^2 - a^2}$$

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: $0.04 = \frac{4}{100} = \frac{1}{25} = 5^{-2}$.
$$(0.04)^{-2.5} = (5^{-2})^{-5/2} = 5^{(-2) \times (-5/2)} = 5^5 = \mathbf{3125}$$

(b): Convert all factors to base 6 ($36 = 6^2, 216 = 6^3$):
$$6^{6.5} \times (6^2)^{4.5} \div (6^3)^{4.5} = 6^{6.5} \times 6^9 \div 6^{13.5} = 6^{6.5 + 9 - 13.5} = 6^2$$
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: $2\log(5/9) = \log(25/81)$.

$$\log\left[\frac{75}{16} \times \frac{81}{25} \times \frac{32}{243}\right] = \log\left[\left(\frac{75}{25}\right) \times \left(\frac{32}{16}\right) \times \left(\frac{81}{243}\right)\right] = \log\left[3 \times 2 \times \frac{1}{3}\right] = \mathbf{\log 2}$$

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

$$\log\left[\frac{225}{324} \times \frac{162}{25} \times \frac{4}{9}\right] = \log\left[9 \times \frac{1}{2} \times \frac{4}{9}\right] = \mathbf{\log 2}$$

STEP 1 Expand the quotient and product using logarithm rules:

$$\log\left(\frac{10x}{y^2}\right) = \log 10 + \log x - \log(y^2) = 1 + \log x - 2\log y$$

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 $\log_{b^k} x = \frac{1}{k}\log_b x$ to convert all terms to base 2:

• $\log_4 x = \log_{2^2} x = \frac{1}{2}\log_2 x$
• $\log_{16} x = \log_{2^4} x = \frac{1}{4}\log_2 x$

STEP 2 Factor out $\log_2 x$:

$$\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{21}{4} \implies \frac{7}{4}\log_2 x = \frac{21}{4}$$

STEP 3 Multiply both sides by $\frac{4}{7}$: $\log_2 x = 3 \implies x = 2^3 = \mathbf{8}$.

(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 $x^r$:

$$= \frac{x^p}{x^p + x^q + x^r} + \frac{x^q}{x^q + x^p + x^r} + \frac{x^r}{x^r + x^p + x^q} = \frac{x^p + x^q + x^r}{x^p + x^q + x^r} = \mathbf{1}$$

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: $12 = 2^2 \times 3$.

Substitute their representations in terms of $k$:

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

STEP 3 Equate exponents on base $k$:

$$\frac{1}{z} = \frac{2}{x} + \frac{1}{y} \implies \frac{2}{x} = \frac{1}{z} - \frac{1}{y} = \frac{y - z}{yz}$$

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

STEP 1 Express all terms in base 2 ($1024 = 2^{10}, 16 = 2^4, 4 = 2^2$):

• Numerator: $(2^{10})^n \times (2^2)^{2n+1} = 2^{10n} \times 2^{4n+2} = 2^{14n+2}$
• Denominator: $(2^4)^n \times (2^2)^{n-1} = 2^{4n} \times 2^{2n-2} = 2^{6n-2}$

STEP 2 Subtract denominator power: $2^{(14n+2) - (6n-2)} = 2^{8n+4} = (2^4)^{2n+1} = \mathbf{16^{2n+1}}$ (or $4^{4n+2}$).

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

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

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: $0.40z = 1.20 \implies z = \frac{1.20}{0.40} = \mathbf{3}$.

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

(ii): Let $K = \sqrt{\frac{(238.2)^2 \times 86.34}{92.3}}$.
$$\log K = \frac{1}{2}[2\log(238.2) + \log(86.34) - \log(92.3)]$$ $$= \frac{1}{2}[2(2.3770) + 1.9362 - 1.9652] = \frac{1}{2}[4.7540 - 0.0290] = \frac{1}{2}[4.7250] = 2.3625$$ $$K = \text{Antilog}(2.3625) = \mathbf{230.4}$$.

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

PART (ii) At year 2020 ($t_1 = 10$), $P_{2020} = 4.7(1.02)^{10}$. We seek year $t$ where $P_t = 2 \times P_{2020}$:
$$4.7(1.02)^t = 2 \times 4.7(1.02)^{10} \implies (1.02)^{t-10} = 2$$ Taking logarithms on both sides:
$$(t - 10)\log(1.02) = \log 2 \implies t - 10 = \frac{\log 2}{\log 1.02} = \frac{0.30103}{0.00860} \approx 35\text{ years}$$ 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 $13 = 8 + 4 + 1 = 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = (1101)_2$. 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: $\log_{10} x > 0 \implies x > 10^0 \implies x > 1$. Reason R gives the exact domain condition of the logarithmic function.