1st Law (Product Law) $$\log_b(mn) = \log_b m + \log_b n$$

2nd Law (Quotient Law) $$\log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n$$

3rd Law (Power Law) $$\log_b(m^n) = n \log_b m$$

4th Law (Change of Base Law) $$\log_b m = \frac{\log_a m}{\log_a b} \iff \log_b m \cdot \log_a b = \log_a m$$

1. Earthquake Richter Scale Model The magnitude $M$ on the Richter scale is given by: $$M = \log_{10}\left(\frac{I}{I_0}\right)$$ where $I$ is shock wave intensity and $I_0$ is standard threshold intensity.

Intensity Comparison Ratio: If two earthquakes have magnitudes $M_1$ and $M_2$: $$\frac{I_2}{I_1} = 10^{M_2 - M_1} = \text{antilog}(M_2 - M_1)$$ Example: China (1978, $M_2 = 8.2$) vs Pakistan (2005, $M_1 = 7.6$): $$\frac{I_2}{I_1} = 10^{8.2 - 7.6} = 10^{0.6} \approx \mathbf{3.98 \approx 4\text{ times stronger}}.$$

2. Number of Digits in $a^b$ To find the number of digits in a giant power $N = a^b$: $$\log_{10} N = \log_{10}(a^b) = b \log_{10} a = c.m$$ $$\mathbf{\text{Number of Digits} = \text{Characteristic}(c) + 1}$$ Example ($5^{50}$): $$\log_{10}(5^{50}) = 50 \log_{10} 5 = 50(0.6990) = 34.95$$ $$\text{Characteristic} = 34 \implies \text{Total Digits} = 34 + 1 = \mathbf{35\text{ digits}}.$$

Part (i) $5700$

• Step 1: Locate original decimal point & reference position:
• In $5700$, the decimal point is at the end: $5700.$
• The first non-zero digit from left is $5$. The reference position is immediately after $5$ ($5_{\wedge}700$).
• Step 2: Count the shift of decimal point:
• We move the decimal point $3$ places to the left to reach the reference position.
• Moving left corresponds to a positive exponent: $n = +3$.
• Step 3: Write in standard scientific notation:

Part (ii) $49,800,000$

• Step 1: Locate original decimal point & reference position:
• In $49,800,000$, the decimal point is at the end: $49800000.$
• The reference position is after the first non-zero digit $4$ ($4_{\wedge}9800000$).
• Step 2: Count decimal shift:
• Move the decimal point $7$ places to the left: $n = +7$.
• Step 3: Formulate result:

Part (iii) $0.0000000016$

• Step 1: Locate original decimal point & reference position:
• The number is smaller than 1: $0.0000000016$.
• The first non-zero digit is $1$. The reference position is after $1$ ($1_{\wedge}6$).
• Step 2: Count decimal shift:
• Move the decimal point $9$ places to the right to reach after $1$: $n = -9$.
• Step 3: Formulate result:

Part (iv) $0.0074$

• Step 1: Locate original decimal point & reference position:
• The number is $0.0074$. First non-zero digit is $7$ ($7_{\wedge}4$).
• Step 2: Count decimal shift:
• Move the decimal point $3$ places to the right: $n = -3$.
• Step 3: Formulate result:

Part (i) $6.35 \times 10^7$

• Analysis: The exponent $+7$ is positive, so we shift the decimal point $7$ places to the right.
• Step-by-step expansion:

Part (ii) $7.61 \times 10^{-4}$

• Analysis: The exponent $-4$ is negative, so we shift the decimal point $4$ places to the left.
• Step-by-step expansion:

Part (iii) $9.018 \times 10^{-6}$

• Analysis: The exponent $-6$ is negative, so we shift the decimal point $6$ places to the left.
• Step-by-step expansion:

Part (iv) $5.06 \times 10^8$

• Analysis: The exponent $+8$ is positive, so we shift the decimal point $8$ places to the right.
• Step-by-step expansion:

Part (i) $(3 \times 10^4) \times (2 \times 10^5)$

• Step 1: Group coefficients and powers of 10:

• Step 2: Multiply coefficients and add exponents ($a^m \cdot a^n = a^{m+n}$):

Part (ii) $\frac{8.4 \times 10^8}{2.1 \times 10^3}$

• Step 1: Group numerical factors and exponential factors:

• Step 2: Divide coefficients and subtract exponents ($\frac{a^m}{a^n} = a^{m-n}$):

Part (iii) $\frac{(4 \times 10^{-3}) \times (6 \times 10^7)}{2 \times 10^2}$

• Step 1: Simplify numerator first:

• Step 2: Divide by denominator:

• Step 3: Convert to standard scientific notation ($1 \le A < 10$):

Part (iv) $\frac{1.2 \times 10^{-5}}{3 \times 10^{-8}}$

• Step 1: Separate coefficients and powers:

• Step 2: Apply quotient rule of exponents:

• Step 3: Adjust to scientific notation:

> Problem Statement: If it takes $5\text{ seconds}$ to recite 'Kalma Pak' once, how many hours will it take to recite 'Kalma Pak' one million ($10^6$) times? Express the answer in standard and scientific notation.

• Given Data:
• Time for 1 recitation = $5\text{ seconds}$
• Total number of recitations = $1,000,000 = 10^6$
• Step 1: Calculate total time in seconds:

• Step 2: Convert seconds into hours:
• Since $1\text{ hour} = 60 \times 60 = 3600\text{ seconds}$:

• Step 3: Express in standard and scientific notation:
• Standard Decimal Form: $\mathbf{1388.89\text{ hours}}$ (or $57\text{ days } 20\text{ hours } 53\text{ minutes } 20\text{ seconds}$)
• Scientific Notation: $\mathbf{1.3889 \times 10^3\text{ hours}}$

> Problem Statement: The distance between the Earth and the Sun is approximately $1.5 \times 10^8\text{ km}$. The speed of light is $3.0 \times 10^5\text{ km/s}$. Find the time taken by sunlight to reach the Earth in seconds, and convert it into minutes and seconds.

• Given Data:
• Distance ($S$) = $1.5 \times 10^8\text{ km}$
• Speed of light ($v$) = $3.0 \times 10^5\text{ km/s}$
• Step 1: Apply formula $S = v \times t \implies t = \frac{S}{v}$:

• Step 2: Convert total seconds into minutes and remaining seconds:

• Final Result:
• Time in seconds: $\mathbf{500\text{ s}} = \mathbf{5 \times 10^2\text{ s}}$
• Time in minutes and seconds: $\mathbf{8\text{ minutes } 20\text{ seconds}}$

Part (i) $\log_3(-1)$

• Analysis:
• For $\log_b x$ to be defined in $\mathbb{R}$, the argument must be strictly positive ($x > 0$) and the base must satisfy $b > 0, b \ne 1$.
• Here, base $b = 3 > 0, 3 \ne 1$, but argument $x = -1 \le 0$.
• If we let $\log_3(-1) = y$, then $3^y = -1$. Since any real power of a positive base $3$ is strictly positive ($3^y > 0$ for all $y \in \mathbb{R}$), no real power can produce a negative number.
• Conclusion:

Part (ii) $\log_2(0)$

• Analysis:
• The argument is $x = 0$.
• If $\log_2 0 = y \implies 2^y = 0$.
• As $y \to -\infty$, $2^y \to 0$, but $2^y$ is never equal to $0$ for any finite real value of $y$.
• Conclusion:

Part (iii) $\log_2(4 - 2x)$ for specific values of $x$:

• General Domain Requirement:

• Case 1: At $x = 1$:
• Argument $= 4 - 2(1) = 4 - 2 = 2 > 0$.
• Expression $= \log_2(2) = 1$.
• Result: $\mathbf{\text{DEFINED, value } = 1}$.

• Case 2: At $x = -1$:
• Argument $= 4 - 2(-1) = 4 + 2 = 6 > 0$.
• Expression $= \log_2(6)$.
• Result: $\mathbf{\text{DEFINED}}$.

• Case 3: At $x = 6$:
• Argument $= 4 - 2(6) = 4 - 12 = -8 < 0$.
• Expression $= \log_2(-8)$.
• Result: $\mathbf{\text{UNDEFINED}}$ (logarithm of a negative number does not exist in $\mathbb{R}$).

> Fundamental Rule: $b^y = x \iff \log_b x = y$ (where $b > 0, b \ne 1, x > 0$).

| Part | Given Form | Conversion Process | Converted Equivalent Form |

| : | : | : | : |

| (i) | $a^y = x$ (Exponential) | Base is $a$, exponent is $y$, result is $x$ | $\mathbf{\log_a x = y}$ |

| (ii) | $2^3 = 8$ (Exponential) | Base is $2$, exponent is $3$, result is $8$ | $\mathbf{\log_2 8 = 3}$ |

| (iii) | $3^4 = 81$ (Exponential) | Base is $3$, exponent is $4$, result is $81$ | $\mathbf{\log_3 81 = 4}$ |

| (iv) | $10^{-2} = 0.01$ (Exponential) | Base is $10$, exponent is $-2$, result is $0.01$ | $\mathbf{\log_{10} 0.01 = -2}$ |

| (v) | $5^0 = 1$ (Exponential) | Base is $5$, exponent is $0$, result is $1$ | $\mathbf{\log_5 1 = 0}$ |

| (vi) | $\log_2 16 = 4$ (Logarithmic) | Base is $2$, power is $4$, value is $16$ | $\mathbf{2^4 = 16}$ |

| (vii) | $\log_{10} 0.001 = -3$ (Logarithmic) | Base is $10$, power is $-3$, value is $0.001$ | $\mathbf{10^{-3} = 0.001}$ |

| (viii) | $\log_3 \left(\frac{1}{9}\right) = -2$ (Logarithmic) | Base is $3$, power is $-2$, value is $\frac{1}{9}$ | $\mathbf{3^{-2} = \frac{1}{9}}$ |

| (ix) | $\ln x = y \iff \log_e x = y$ (Logarithmic) | Base is $e$, power is $y$, value is $x$ | $\mathbf{e^y = x}$ |

| (x) | $\log_a 1 = 0$ (Logarithmic) | Base is $a$, power is $0$, value is $1$ | $\mathbf{a^0 = 1}$ |

Part (i) $\log_3 x = 4$

• Step 1: Convert to exponential form:

• Step 2: Evaluate:

Part (ii) $\log_x 64 = 2$

• Step 1: Convert to exponential form:

• Step 2: Solve for $x$ (taking principal positive square root since base $x > 0$):

Part (iii) $\log_4 16 = x$

• Step 1: Convert to exponential form:

• Step 2: Express both sides with base $4$:

Part (iv) $\log_x 0.001 = -3$

• Step 1: Convert to exponential form:

• Step 2: Express decimal as fraction & negative power:

• Step 3: Equate bases since exponents are equal:

Part (v) $\log_{10} 1000 = x$

• Step 1: Convert to exponential form:

• Step 2: Express $1000$ as power of $10$:

Part (vi) $\log_2 x = -3$

• Step 1: Convert to exponential form:

• Step 2: Evaluate using negative exponent rule:

Part (i) $\log_{81} 9 = x$

• Convert to exponential form: $81^x = 9$
• Express with common base $9$: $(9^2)^x = 9^1 \implies 9^{2x} = 9^1$
• Equate exponents: $2x = 1 \implies \mathbf{x = \frac{1}{2}}$

Part (ii) $\log_a 6 = 0.5$

• Convert to exponential form: $a^{0.5} = 6 \implies a^{1/2} = 6$
• Square both sides: $(a^{1/2})^2 = 6^2 \implies \mathbf{a = 36}$

Part (iii) $\log_5 n = 2$

• Convert to exponential form: $n = 5^2 \implies \mathbf{n = 25}$

Part (iv) $10^p = 40$

• Convert to logarithmic form: $\mathbf{p = \log_{10} 40 \approx 1.6021}$

| # | Number $N$ | Reference Position & Shift | Characteristic ($c$) | Mantissa Calculation ($m$) | Final Logarithm $\log_{10} N$ |

| :: | : | : | :: | : | :: |

| 1 | $5313$ | $5_{\wedge}313.$ ($3$ digits left of ref) | $+3$ | Row $53$, Col $1 \to 7251$; Mean diff Col $3 \to 2$.
$m = 7251 + 2 = .7253$ | $\mathbf{3.7253}$ |

| 2 | $4580$ | $4_{\wedge}580.$ ($3$ digits left of ref) | $+3$ | Row $45$, Col $8 \to 6609$; Mean diff Col $0 \to 0$.
$m = .6609$ | $\mathbf{3.6609}$ |

| 3 | $9.613$ | $9_{\wedge}.613$ ($0$ digits shift) | $0$ | Row $96$, Col $1 \to 9827$; Mean diff Col $3 \to 1$.
$m = 9827 + 1 = .9828$ | $\mathbf{0.9828}$ |

| 4 | $110.9$ | $1_{\wedge}10.9$ ($2$ digits left of ref) | $+2$ | Row $11$, Col $0 \to 0414$; Mean diff Col $9 \to 35$.
$m = 0414 + 35 = .0449$ | $\mathbf{2.0449}$ |

| 5 | $52.39$ | $5_{\wedge}2.39$ ($1$ digit left of ref) | $+1$ | Row $52$, Col $3 \to 7185$; Mean diff Col $9 \to 7$.
$m = 7185 + 7 = .7192$ | $\mathbf{1.7192}$ |

| 6 | $0.01207$ | $0.01_{\wedge}207$ ($1$ zero after decimal) | $-2 = \bar{2}$ | Row $12$, Col $0 \to 0792$; Mean diff Col $7 \to 25$.
$m = 0792 + 25 = .0817$ | $\mathbf{\bar{2}.0817}$ ($-1.9183$) |

| 7 | $0.0093$ | $0.009_{\wedge}3$ ($2$ zeros after decimal) | $-3 = \bar{3}$ | Row $93$, Col $0 \to 9685$; Mean diff $0 \to 0$.
$m = .9685$ | $\mathbf{\bar{3}.9685}$ ($-2.0315$) |

| 8 | $1\text{ Trillion} = 10^{12}$ | $1.0 \times 10^{12}$ | $+12$ | $\log_{10}(10^{12}) = 12 \log_{10}(10) = 12(1) = 12.0000$ | $\mathbf{12.0000}$ |

| 9 | $0.00004$ | $0.00004_{\wedge}$ ($4$ zeros after decimal) | $-5 = \bar{5}$ | Row $40$, Col $0 \to 6021$.
$m = .6021$ | $\mathbf{\bar{5}.6021}$ ($-4.3979$) |

| 10 | $4$ | $4_{\wedge}.000$ ($1$ digit, $c = 1-1=0$) | $0$ | Row $40$, Col $0 \to 6021$.
$m = .6021$ | $\mathbf{0.6021}$ |

| 11 | $4000$ | $4_{\wedge}000.$ ($4$ digits, $c = 4-1=3$) | $+3$ | Row $40$, Col $0 \to 6021$.
$m = .6021$ | $\mathbf{3.6021}$ |

| 12 | $54$ | $5_{\wedge}4.$ ($2$ digits, $c = 2-1=1$) | $+1$ | Row $54$, Col $0 \to 7324$.
$m = .7324$ | $\mathbf{1.7324}$ |

| 13 | $2.15$ | $2_{\wedge}.15$ ($1$ digit, $c = 1-1=0$) | $0$ | Row $21$, Col $5 \to 3324$.
$m = .3324$ | $\mathbf{0.3324}$ |

| 14 | $2^{15}$ | $2^{15} = 32768$ | $+4$ | $\log(2^{15}) = 15 \log(2) = 15(0.30103) = 4.51545$.
Table: Row $32$, Col $7 \to 5145$; diff $6 \to 8 \implies .5154$ | $\mathbf{4.5154}$ |

| 15 | $-4$ | Negative argument ($x = -4 \le 0$) | N/A | $\log_{10}(-4)$ has no real value because $10^y > 0$ for all $y \in \mathbb{R}$. | $\mathbf{\text{UNDEFINED in } \mathbb{R}}$ |

## 📝 Exercise 2.4 — Antilogarithms Step-by-Step Solutions

> Methodology for Finding Antilogarithm:

> Given $y = c.m$ (where $c$ is the characteristic and $m$ is the mantissa $\ge 0$):

> 1. Table Reading using Mantissa $m$: Look up the first two decimal digits of $m$ in the left-most column of the Antilogarithm Table, find the value under the 3rd decimal digit column, and add the mean difference under the 4th decimal digit column to get a 4-digit sequence.

> 2. Place Reference Position: Mark the reference position immediately after the first non-zero digit of the sequence ($d_1{}_{\wedge}d_2 d_3 d_4$).

> 3. Locate Decimal Point using Characteristic $c$:

> * If $c \ge 0$: Shift decimal point $c$ places to the right from the reference position.

> * If $c = \bar{p} < 0$: Shift decimal point $p$ places to the left from the reference position (placing $p-1$ zeros between decimal point and first non-zero digit).

| # | Given Log Value $y$ | Characteristic ($c$) & Mantissa ($m$) | Antilog Table Lookup Sequence | Decimal Point Placement Rule | Final Antilogarithm Value $\text{antilog}(y)$ |

| :: | : | : | : | : | :: |

| 1 | $2.4324$ | $c = 2$, $m = .4324$ | Row $.43$, Col $2 \to 2704$; Diff $4 \to 3$.
Seq: $2704 + 3 = 2707$ | Ref: $2_{\wedge}707$. Shift right $2$ places: $270.7$ | $\mathbf{270.7}$ |

| 2 | $1.5890$ | $c = 1$, $m = .5890$ | Row $.58$, Col $9 \to 3882$; Diff $0 \to 0$.
Seq: $3882$ | Ref: $3_{\wedge}882$. Shift right $1$ place: $38.82$ | $\mathbf{38.82}$ |

| 3 | $0.2425$ | $c = 0$, $m = .2425$ | Row $.24$, Col $2 \to 1746$; Diff $5 \to 2$.
Seq: $1746 + 2 = 1748$ | Ref: $1_{\wedge}748$. Shift $0$ places: $1.748$ | $\mathbf{1.748}$ |

| 4 | $3.5636$ | $c = 3$, $m = .5636$ | Row $.56$, Col $3 \to 3656$; Diff $6 \to 5$.
Seq: $3656 + 5 = 3661$ | Ref: $3_{\wedge}661$. Shift right $3$ places: $3661.$ | $\mathbf{3661}$ |

| 5 | $0.0038$ | $c = 0$, $m = .0038$ | Row $.00$, Col $3 \to 1007$; Diff $8 \to 2$.
Seq: $1007 + 2 = 1009$ | Ref: $1_{\wedge}009$. Shift $0$ places: $1.009$ | $\mathbf{1.009}$ |

| 6 | $0.0000$ | $c = 0$, $m = .0000$ | Row $.00$, Col $0 \to 1000$.
Seq: $1000$ | Ref: $1_{\wedge}000$. Shift $0$ places: $1.000 = 10^0$ | $\mathbf{1.000}$ |

| 7 | $\bar{1}.2429$ | $c = \bar{1} = -1$, $m = .2429$ | Row $.24$, Col $2 \to 1746$; Diff $9 \to 4$.
Seq: $1746 + 4 = 1750$ | Ref: $1_{\wedge}750$. Shift left $1$ place: $0.1750$ | $\mathbf{0.1750}$ |

| 8 | $\bar{1}.9281$ | $c = \bar{1} = -1$, $m = .9281$ | Row $.92$, Col $8 \to 8472$; Diff $1 \to 2$.
Seq: $8472 + 2 = 8474$ | Ref: $8_{\wedge}474$. Shift left $1$ place: $0.8474$ | $\mathbf{0.8474}$ |

| 9 | $1.5219$ | $c = 1$, $m = .5219$ | Row $.52$, Col $1 \to 3319$; Diff $9 \to 7$.
Seq: $3319 + 7 = 3326$ | Ref: $3_{\wedge}326$. Shift right $1$ place: $33.26$ | $\mathbf{33.26}$ |

| 10 | $0.4900$ | $c = 0$, $m = .4900$ | Row $.49$, Col $0 \to 3090$; Diff $0 \to 0$.
Seq: $3090$ | Ref: $3_{\wedge}090$. Shift $0$ places: $3.090$ | $\mathbf{3.090}$ |

| 11 | $\bar{2}.4900$ | $c = \bar{2} = -2$, $m = .4900$ | Row $.49$, Col $0 \to 3090$.
Seq: $3090$ | Ref: $3_{\wedge}090$. Shift left $2$ places: $0.03090$ | $\mathbf{0.03090}$ |

| 12 | $5.9990$ | $c = 5$, $m = .9990$ | Row $.99$, Col $9 \to 9977$; Diff $0 \to 0$.
Seq: $9977$ | Ref: $9_{\wedge}977$. Shift right $5$ places: $997,700$ | $\mathbf{997,700}$ |

| 13 | $3.4900$ | $c = 3$, $m = .4900$ | Row $.49$, Col $0 \to 3090$.
Seq: $3090$ | Ref: $3_{\wedge}090$. Shift right $3$ places: $3090.$ | $\mathbf{3090}$ |

| 14 | $-3 = \bar{3}.0000$ | $c = \bar{3} = -3$, $m = .0000$ | Row $.00$, Col $0 \to 1000$.
Seq: $1000$ | Ref: $1_{\wedge}000$. Shift left $3$ places: $0.001000 = 10^{-3}$ | $\mathbf{0.001}$ |

| 15 | $2.34 = 2.3400$ | $c = 2$, $m = .3400$ | Row $.34$, Col $0 \to 2188$.
Seq: $2188$ | Ref: $2_{\wedge}188$. Shift right $2$ places: $218.8$ | $\mathbf{218.8}$ |

Part (i) $\log \left(\frac{9\sqrt{t}}{59}\right)$

• Step 1: Apply quotient law:

• Step 2: Apply product law on numerator:

• Step 3: Apply power law on radical ($\sqrt{t} = t^{1/2}$):

Part (ii) $\log \left(\frac{52 t^5 a^3}{s(5pq^2)}\right)$

• Step 1: Apply quotient law:

• Step 2: Apply product law on numerator and denominator:

• Step 3: Apply power law ($n \log x$) and distribute negative sign:

Part (iii) $\log \left(\frac{\sqrt[3]{x^2 y}}{\sqrt{z}}\right)$

• Step 1: Convert radicals to fractional exponents:

• Step 2: Apply quotient and product laws:

• Step 3: Apply power law:

Part (iv) $\log \left(\frac{46.4}{\sqrt{53.3}}\right)$

• Step 1: Apply quotient law:

• Step 2: Apply power law on radical ($(53.3)^{1/2}$):

Part (v) $\log \sqrt{\frac{d^6 b^2}{4.4 t b^3}}$

• Step 1: Simplify algebraic fraction inside radical:

• Step 2: Apply power law for outer radical ($\frac{1}{2}$):

• Step 3: Expand numerator and denominator:

Part (vi) $\log \left(\frac{x^3 \sqrt{y}}{z^4}\right)$

• Step 1: Apply quotient law:

• Step 2: Expand using product and power laws:

Part (vii) $\log \left(\frac{7^2 \sqrt{t^3 p}}{53.3}\right)$

• Step 1: Express radical with rational powers ($(t^3 p)^{1/2} = t^{3/2} p^{1/2}$):

• Step 2: Apply power law:

Part (i) $3\log x - 5\log y$

• Step 1: Apply power law backwards ($n\log a = \log a^n$):

• Step 2: Apply quotient law backwards ($\log a - \log b = \log \frac{a}{b}$):

Part (ii) $\frac{1}{2}\log t - \log r - \frac{1}{5}\log s$

• Step 1: Group positive and negative terms:

• Step 2: Combine inside brackets via product law:

• Step 3: Combine via quotient law:

Part (iii) $\frac{1}{3}[\log 57.7 + 4\log 36.6 - 2\log 23.3]$

• Step 1: Simplify terms inside square brackets:

• Step 2: Apply outer coefficient $\frac{1}{3}$ as cube root:

Part (iv) $5\log 6 - 7\log 9.42 + \frac{1}{2}\log t - \frac{1}{3}\log 32.2 + \frac{3}{4}\log a$

• Step 1: Separate into positive terms (numerator) and negative terms (denominator):

• Step 2: Combine into single logarithm:

Part (v) $\frac{1}{4}\log 37.74 - \frac{5}{4}\log 53.11 - \frac{1}{2}\log 28.83$

• Step 1: Write terms as powers:

• Step 2: Combine into a single quotient term:

Part (i) $\log_7 343$

• Step 1: Express $343$ as a power of $7$: $343 = 7^3$.
• Step 2: Apply power law and identity $\log_b b = 1$:

Part (ii) $\log \left(\frac{72.34}{15}\right)$

• Step 1: Apply quotient law:

• Step 2: Find log values using log tables:
• $\log(72.34): c = 1, m = .8594 \implies \log(72.34) = 1.8594$
• $\log(15): c = 1, m = .1761 \implies \log(15) = 1.1761$
• Step 3: Subtract:

Part (iii) $\log \sqrt[5]{65}$

• Step 1: Apply power law:

• Step 2: Find $\log(65)$ from table:
• $\log(65): c = 1, m = .8129 \implies \log(65) = 1.8129$
• Step 3: Multiply by $\frac{1}{5}$:

Part (i) $\log_b \left(\frac{6}{5}\right)$

• Step 1: Decompose numerator into prime factors: $6 = 2 \times 3$.
• Step 2: Apply quotient and product laws:

• Step 3: Substitute given values:

Part (ii) $\log_b \sqrt{\frac{2}{9}}$

• Step 1: Express with prime factors and fractional exponent:

• Step 2: Apply power and quotient laws:

• Step 3: Substitute values:

• Step 4: Convert negative decimal into characteristic & mantissa:

Part (iii) $\log_b \left(\frac{45(3)^2}{100}\right)$

• Step 1: Simplify numerical fraction:

• Step 2: Expand using laws of logarithms:

• Step 3: Substitute given values:

Part (iv) $\log_b 0.024$

• Step 1: Convert decimal to prime factorized fraction:

• Step 2: Expand:

• Step 3: Substitute values:

• Step 4: Convert into standard notation ($\bar{c}.m$):

Part (v) $\log_b \sqrt[5]{5^2}$

• Step 1: Express in exponential form: $\sqrt[5]{5^2} = 5^{2/5}$.
• Step 2: Apply power law:

• Step 3: Substitute $\log_b 5 = 0.6990$:

> Mathematical Principle:

> For any positive integer $N$, the number of digits is given by:

> $$\text{Number of Digits in } N = \text{Characteristic of } \log_{10} N + 1 = \lfloor \log_{10} N \rfloor + 1$$

Part (i) $2^{10}$

• Step 1: Let $x = 2^{10}$ and take common logarithm:

• Step 2: Substitute $\log 2 = 0.3010$:

• Step 3: Extract characteristic and compute digits:

Part (ii) $3^{20}$

• Step 1: Let $x = 3^{20}$ and take common logarithm:

• Step 2: Substitute $\log 3 = 0.4771$:

• Step 3: Extract characteristic and compute digits:

Part (iii) $5^{50}$

• Step 1: Let $x = 5^{50}$ and take common logarithm:

• Step 2: Substitute $\log 5 = 0.6990$:

• Step 3: Extract characteristic and compute digits:

Part (iv) $5^{20}$

• Step 1: Let $x = 5^{20}$ and take common logarithm:

• Step 2: Substitute $\log 5 = 0.6990$:

• Step 3: Extract characteristic and compute digits:

Part (v) $529^{30}$

• Step 1: Let $x = 529^{30}$ and take common logarithm:

• Step 2: Find $\log(529)$ from log tables:
• For $529$: $c = 2$, mantissa from row $52$, col $9 = .7235 \implies \log 529 = 2.7235$.
• Step 3: Multiply:

• Step 4: Extract characteristic and compute digits:

Part (i) $23.57 \times 5.967$

• Step 1: Let $x = 23.57 \times 5.967$:
• Step 2: Take common log on both sides:

• Step 3: Find log values from table:
• $\log 23.57: c = 1$, row $23$, col $5 \to 3711$, diff $7 \to 13 \implies m = .3724 \implies \log 23.57 = 1.3724$
• $\log 5.967: c = 0$, row $59$, col $6 \to 7752$, diff $7 \to 5 \implies m = .7757 \implies \log 5.967 = 0.7757$
• Step 4: Add logarithms:

• Step 5: Take antilogarithm on both sides:

• For mantissa $.1481$: row $.14$, col $8 \to 1406$, diff $1 \to 0 \implies 1406$.
• Characteristic $c = 2 \implies$ shift decimal $2$ places right from ref $1_{\wedge}406 \implies 140.6$.
• Final Result:

Part (ii) $\frac{47.27}{9.71 \times 4.171}$

• Step 1: Let $x = \frac{47.27}{9.71 \times 4.171}$:
• Step 2: Take log on both sides:

• Step 3: Look up individual logarithms:
• $\log 47.27: c = 1$, row $47$, col $2 \to 6739$, diff $7 \to 6 \implies 1.6745$
• $\log 9.71: c = 0$, row $97$, col $1 \to 9872 \implies 0.9872$
• $\log 4.171: c = 0$, row $41$, col $7 \to 6201$, diff $1 \to 1 \implies 0.6202$
• Step 4: Perform arithmetic:

• Step 5: Take antilogarithm:

• Mantissa $.0671 \implies$ row $.06$, col $7 \to 1167$, diff $1 \to 0 \implies 1167$.
• Characteristic $c = 0 \implies 1.167$.
• Final Result:

Part (iii) $\frac{5.321 \times 65.89}{7392}$

• Step 1: Let $x = \frac{5.321 \times 65.89}{7392}$:
• Step 2: Take common log:

• Step 3: Table lookups:
• $\log 5.321: c = 0$, row $53$, col $2 \to 7259$, diff $1 \to 1 \implies 0.7260$
• $\log 65.89: c = 1$, row $65$, col $8 \to 8182$, diff $9 \to 6 \implies 1.8188$
• $\log 7392: c = 3$, row $73$, col $9 \to 8686$, diff $2 \to 1 \implies 3.8687$
• Step 4: Compute $\log x$:

• Step 5: Convert negative result into characteristic & mantissa:

• Step 6: Take antilogarithm:

• Mantissa $.6761 \implies$ row $.67$, col $6 \to 4742$, diff $1 \to 1 \implies 4743$.
• Characteristic $c = \bar{2} \implies$ shift $2$ places left: $0.04743$.
• Final Result:

Part (iv) $\frac{\sqrt{28.73}}{129.4}$

• Step 1: Let $x = \frac{(28.73)^{1/2}}{129.4}$:
• Step 2: Take logarithm:

• Step 3: Table lookups:
• $\log 28.73: c = 1$, row $28$, col $7 \to 4579$, diff $3 \to 5 \implies 1.4584$
• $\frac{1}{2}(1.4584) = 0.7292$
• $\log 129.4: c = 2$, row $12$, col $9 \to 1106$, diff $4 \to 14 \implies 2.1120$
• Step 4: Subtract:

• Step 5: Take antilogarithm:

• Mantissa $.6172 \implies$ row $.61$, col $7 \to 4140$, diff $2 \to 2 \implies 4142$.
• Characteristic $c = \bar{2} \implies 0.04142$.
• Final Result:

Part (v) $\sqrt[3]{28.73}$

• Step 1: Let $x = (28.73)^{1/3}$:
• Step 2: Take log:

• Step 3: Take antilogarithm:

• Mantissa $.4861 \implies$ row $.48$, col $6 \to 3062$, diff $1 \to 1 \implies 3063$.
• Characteristic $c = 0 \implies 3.063$.
• Final Result:

Part (vi) $(28.73)^3 \times 27.98$

• Step 1: Let $x = (28.73)^3 \times 27.98$:
• Step 2: Take log:

• Step 3: Evaluate individual terms:
• $\log 28.73 = 1.4584 \implies 3 \times 1.4584 = 4.3752$
• $\log 27.98: c = 1$, row $27$, col $9 \to 4456$, diff $8 \to 13 \implies 1.4469$
• Step 4: Add:

• Step 5: Take antilogarithm:

• Mantissa $.8221 \implies$ row $.82$, col $2 \to 6637$, diff $1 \to 2 \implies 6639$.
• Characteristic $c = 5 \implies 663,900$.
• Final Result:

Part (vii) $\sqrt[3]{16} \times \sqrt[4]{238}$

• Step 1: Let $x = 16^{1/3} \times 238^{1/4}$:
• Step 2: Take logarithm:

• Step 3: Table lookups:
• $\log 16: c = 1$, row $16$, col $0 \to 2041 \implies 1.2041 \implies \frac{1.2041}{3} = 0.4014$
• $\log 238: c = 2$, row $23$, col $8 \to 3766 \implies 2.3766 \implies \frac{2.3766}{4} = 0.5942$
• Step 4: Add:

• Step 5: Take antilogarithm:

• Mantissa $.9956 \implies$ row $.99$, col $5 \to 9886$, diff $6 \to 14 \implies 9900$.
• Characteristic $c = 0 \implies 9.900$.
• Final Result:

Part (viii) $\frac{(28.73)^3}{129.4}$

• Step 1: Let $x = \frac{(28.73)^3}{129.4}$:
• Step 2: Take logarithm:

• Step 3: Substitute known logarithms:
• $3\log(28.73) = 3(1.4584) = 4.3752$
• $\log(129.4) = 2.1120$
• Step 4: Subtract:

• Step 5: Take antilogarithm:

• Mantissa $.2632 \implies$ row $.26$, col $3 \to 1832$, diff $2 \to 1 \implies 1833$.
• Characteristic $c = 2 \implies 183.3$.
• Final Result:

> Problem Statement: The Kansu, China earthquake of 1920 was measured about $8.5$ on the Richter scale, and the Tokyo, Japan earthquake of 1923 was measured $7.8$ on that scale. How many times stronger was the 1920 Kansu earthquake than the 1923 Tokyo earthquake?

• Given Data:
• Magnitude of Kansu earthquake ($M_1$) = $8.5$
• Magnitude of Tokyo earthquake ($M_2$) = $7.8$
• Richter scale formula: $M = \log_{10}\left(\frac{I}{I_0}\right)$, where $I$ is seismic intensity and $I_0$ is baseline reference intensity.
• Step 1: Formulate equations for both earthquakes:

• Step 2: Subtract Equation 2 from Equation 1:

• Step 3: Convert to exponential form (or take antilogarithm):

• From antilog table: row $.70$, col $0 \to 5012$, diff $0 \to 0$.
• Characteristic is $0 \implies 5.012$.
• Conclusion:

Part (i) If $a = b \times 10^n$ is written in scientific notation, then:

• Options: (a) $0 \le b < 10$ \quad (b) $0 \le b \le 10$ \quad (c) $1 \le b < 10$ \quad (d) $1 \le b \le 10$
• Rationale: By mathematical definition of scientific notation, the mantissa coefficient $b$ must be a real number such that $1 \le b < 10$, and $n$ must be an integer.
• Correct Answer: $\mathbf{(c) \; 1 \le b < 10}$

Part (ii) In $0.537$, the reference position is:

• Options: (a) after $0$ \quad (b) after $7$ \quad (c) after $5$ \quad (d) before $7$
• Rationale: The reference position is the place immediately following the first non-zero digit from the left. In $0.537$, the first non-zero digit is $5$, so the reference position is $0.5_{\wedge}37$.
• Correct Answer: $\mathbf{(c) \; \text{after } 5}$

Part (iii) $\log_2(-2)$ is:

• Options: (a) $-2$ \quad (b) $2$ \quad (c) $0$ \quad (d) impossible
• Rationale: The domain of logarithmic function $\log_b x$ requires $x > 0$. Since $-2 < 0$, no real number $y$ satisfies $2^y = -2$.
• Correct Answer: $\mathbf{(d) \; \text{impossible (undefined in } \mathbb{R})}$

Part (iv) If $\log(x + 3) = \log(15x - 4)$, then $x$ is:

• Options: (a) $2$ \quad (b) $\frac{1}{2}$ \quad (c) $-\frac{1}{2}$ \quad (d) $4$
• Rationale: Since logarithmic function is strictly one-to-one:

• Correct Answer: $\mathbf{(b) \; \frac{1}{2}}$

Part (v) $\log_{13} 13 + \log_{15} 1$ is:

• Options: (a) $13$ \quad (b) $1$ \quad (c) $0$ \quad (d) $15$
• Rationale: By standard logarithmic properties, $\log_b b = 1$ and $\log_b 1 = 0$.

• Correct Answer: $\mathbf{(b) \; 1}$

Part (vi) For $a > 0$, $\log_a a^{-2}$ is:

• Options: (a) $-2$ \quad (b) $2$ \quad (c) $a$ \quad (d) $\frac{1}{a^2}$
• Rationale: Applying power law $\log_b(m^n) = n\log_b m$:

• Correct Answer: $\mathbf{(a) \; -2}$

Part (vii) $\log_b(M + N)$ is:

• Options: (a) $\log_b(MN)$ \quad (b) $\log_b M + \log_b N$ \quad (c) both a and b \quad (d) none of these
• Rationale: Logarithm does not distribute over addition. $\log_b(M + N) \ne \log_b M + \log_b N$.
• Correct Answer: $\mathbf{(d) \; \text{none of these}}$

Part (viii) $\log_b(g^h)$ is:

• Options: (a) $g \log_b h$ \quad (b) $\log_b(gh)$ \quad (c) $(\log_b g) \times h$ \quad (d) $h \log_b g$
• Rationale: By the third fundamental law of logarithms (Power Law), $\log_b(g^h) = h \log_b g$.
• Correct Answer: $\mathbf{(d) \; h \log_b g}$

Part (ix) $\log_b M - \log_b N$ is:

• Options: (a) $\log_b(M - N)$ \quad (b) $\log_b\left(\frac{M}{N}\right)$ \quad (c) $\frac{\log_b M}{\log_b N}$ \quad (d) $\log_b(MN)$
• Rationale: By the Quotient Law of Logarithms, the difference of logarithms is the logarithm of the quotient.
• Correct Answer: $\mathbf{(b) \; \log_b\left(\frac{M}{N}\right)}$

Part (x) $4\log_{\sqrt{10}} 100^2$ is:

• Options: (a) $16$ \quad (b) $32$ \quad (c) $8$ \quad (d) $4$
• Rationale:
• Base $= \sqrt{10} = 10^{1/2}$.
• Argument $= 100^2 = (10^2)^2 = 10^4$.
• By change of base / power law: $\log_{10^{1/2}}(10^4) = \frac{4}{1/2}\log_{10} 10 = 8(1) = 8$.
• Multiply by outer coefficient $4$: $4 \times 8 = 32$.
• Correct Answer: $\mathbf{(b) \; 32}$

Part (xi) $\log 18$ is equal to:

• Options: (a) $3\log 2 + \log 3$ \quad (b) $\log 2 + 2\log 3$ \quad (c) $3\log 3 + 2\log 2$ \quad (d) $2\log 3 + 3\log 2$
• Rationale:

• Correct Answer: $\mathbf{(b) \; \log 2 + 2\log 3}$

Part (xii) $\log 5 - \log 8 + \log 3 - \log 2$ is equal to:

• Options: (a) $\log\left(\frac{5 \times 2}{8 \times 3}\right)$ \quad (b) $\log\left(\frac{15}{16}\right)$ \quad (c) $\log\left(\frac{5}{8}\right)$ \quad (d) $\log\left(\frac{3}{2}\right)$
• Rationale:

• Correct Answer: $\mathbf{(b) \; \log\left(\frac{15}{16}\right)}$

Part (xiii) $\log_{10} 1000$ is:

• Options: (a) $100$ \quad (b) $30$ \quad (c) $3$ \quad (d) $10$
• Rationale: $1000 = 10^3 \implies \log_{10}(10^3) = 3\log_{10} 10 = 3(1) = 3$.
• Correct Answer: $\mathbf{(c) \; 3}$

Part (xiv) Scientific notation of $6.25$ is:

• Options: (a) $6.25 \times 10^1$ \quad (b) $6.25 \times 10^0$ \quad (c) $6.25 \times 10^{-1}$ \quad (d) $0.625 \times 10^2$
• Rationale: Since $6.25$ already satisfies $1 \le 6.25 < 10$, the power of $10$ is zero ($10^0 = 1$).
• Correct Answer: $\mathbf{(b) \; 6.25 \times 10^0}$

Part (xv) The base of natural logarithm is $e$, which is an:

• Options: (a) rational number \quad (b) integer \quad (c) irrational number \quad (d) imaginary number
• Rationale: Euler's number $e \approx 2.718281828459...$ is a transcendental, non-terminating, non-repeating irrational mathematical constant.
• Correct Answer: $\mathbf{(c) \; \text{irrational number}}$

Part (xvi) If $\log_x 25 = 4$, then $x$ is:

• Options: (a) $5$ \quad (b) $\sqrt{5}$ \quad (c) $4$ \quad (d) $25$
• Rationale:

• Correct Answer: $\mathbf{(b) \; \sqrt{5}}$

Part (xvii) $\log_a 10^4 + \log_a 10$ is:

• Options: (a) $\log_a(10^4 - 10)$ \quad (b) $5\log_a 10$ \quad (c) $4\log_a 10$ \quad (d) $3\log_a 10$
• Rationale:

• Correct Answer: $\mathbf{(b) \; 5\log_a 10}$

Part (xviii) $5\log_b 2 - 2\log_b 5$ is:

• Options: (a) $\log_b\left(\frac{32}{25}\right)$ \quad (b) $\log_b 10$ \quad (c) $\log_b\left(\frac{10}{25}\right)$ \quad (d) $\log_b 7$
• Rationale:

• Correct Answer: $\mathbf{(a) \; \log_b\left(\frac{32}{25}\right)}$

• (i) $0.0059$:
• Reference position: $0.005_{\wedge}9$. Decimal moved $3$ places right.
• Result: $\mathbf{5.9 \times 10^{-3}}$
• (ii) $523.4$:
• Reference position: $5_{\wedge}23.4$. Decimal moved $2$ places left.
• Result: $\mathbf{5.234 \times 10^2}$
• (iii) $53.36$:
• Reference position: $5_{\wedge}3.36$. Decimal moved $1$ place left.
• Result: $\mathbf{5.336 \times 10^1}$

• (i) $7232 \times 10^{-2}$:
• Exponent $-2 \implies$ shift decimal $2$ places left: $\mathbf{72.32}$
• (ii) $10.53 \times 10^2$:
• Exponent $+2 \implies$ shift decimal $2$ places right: $\mathbf{1053}$
• (iii) $20.31 \times 10^0$:
• $10^0 = 1 \implies \mathbf{20.31}$

Part (i) $\log_5(5^3) - \log_2(2^3)$

• Step 1: Apply power law: $3\log_5 5 - 3\log_2 2$
• Step 2: Since $\log_5 5 = 1$ and $\log_2 2 = 1$:

Part (ii) $\log_2 4 - \log_3 9$

• Step 1: Express arguments as powers: $\log_2(2^2) - \log_3(3^2)$
• Step 2: Apply identity $\log_b(b^k) = k$:

Part (iii) $\log_b(\log_x x^2 - \log_b b^{-7})$

• Step 1: Evaluate inner terms:
• $\log_x(x^2) = 2\log_x x = 2(1) = 2$
• $\log_b(b^{-7}) = -7\log_b b = -7(1) = -7$
• Step 2: Substitute:

Part (i) $\log_y 9 = \log_y x$

• By one-to-one property of logarithms: $\mathbf{x = 9}$.

Part (ii) $16^{1/x} = 3$

• Take logarithm with base $16$: $\frac{1}{x} = \log_{16} 3 \implies \mathbf{x = \frac{1}{\log_{16} 3} = \log_3 16 \approx 2.5237}$.

Part (iii) $\log_2(x^2 - 1) = \log_2 3$

• Equate arguments: $x^2 - 1 = 3 \implies x^2 = 4 \implies x = \pm 2$.
• Since base and argument must be valid, for $x = 2$ or $x = -2$, $x^2 - 1 = 3 > 0$.
• Thus, $\mathbf{x = 2}$ (or $x = \pm 2$).

Part (iv) $\log_y(8\log_y y) = x$

• Since $\log_y y = 1$, the expression becomes $\log_y(8 \times 1) = x \implies \log_y 8 = x \implies \mathbf{y^x = 8}$.

Part (i) $\log_b 360$

• Prime factorization: $360 = 2^3 \times 3^2 \times 5$.
• Expand: $\log_b(2^3 \times 3^2 \times 5) = 3\log_b 2 + 2\log_b 3 + \log_b 5$.
• Substitute values:

Part (ii) $\log_b 0.24$

• Fraction: $0.24 = \frac{24}{100} = \frac{6}{25} = \frac{2 \times 3}{5^2}$.
• Expand: $\log_b 2 + \log_b 3 - 2\log_b 5$.
• Substitute:

Part (iii) $\log_b 30$

• Prime factorization: $30 = 2 \times 3 \times 5$.
• Expand: $\log_b 2 + \log_b 3 + \log_b 5$.
• Substitute:

• Step 1: Let $x = \frac{0.5327 \times (24.16)^{1/3}}{1.285}$:
• Step 2: Take common logarithm:

• Step 3: Table values:
• $\log 0.5327: c = \bar{1}$, mantissa $.7265 \implies \bar{1}.7265 = -0.2735$
• $\log 24.16: c = 1$, mantissa $.3831 \implies 1.3831 \implies \frac{1.3831}{3} = 0.4610$
• $\log 1.285: c = 0$, mantissa $.1089 \implies 0.1089$
• Step 4: Compute $\log x$:

• Step 5: Take antilogarithm:

• Row $.07$, col $8 \to 1197$, diff $6 \to 2 \implies 1199$.
• Characteristic $c = 0 \implies 1.199$.
• Final Answer:

> Theorem: Prove that $\log_v U \times \log_w V \times \log_u W = 1$ for all valid bases and positive real arguments.

• Proof:
• By the Change of Base Formula, for any convenient common base $10$ (or base $b$):

• Substitute these expressions into the left-hand side ($\text{LHS}$):

• Cancelling common factors across numerators and denominators:

• Conclusion: