00641-MWALA_LEARN-LOGARITHMS-SOLVING-FOREVER-2

Objectives: 0064-MWALA_LEARN-LOGARITHMS-SOLVING-FOREVER-2

Answers β€” Questions 16–20 (Standard Form, Powers & Tables)

Answers β€” Questions 16 to 20

Each question: full steps, symbols explained, a small table where helpful, real-life example, and a short profitable idea.

Question 16

Task: Use mathematical tables to determine the value of (a) (2.09)10 and (b) (0.5216)2/3.

Symbols & methods

Meaning:

  • (2.09)10 means multiply 2.09 by itself 10 times.
  • (0.5216)2/3 means raise 0.5216 to the power 2/3. Equivalent forms: (0.5216)2/3 = ((0.5216)2)1/3 = βˆ›(0.5216Β²). Using logarithm tables one would compute log y = (2/3) log(0.5216) and then find the antilog.

Solution (a) β€” compute (2.09)10

To avoid doing ten successive multiplications, use repeated squaring (table shown). That is: compute powers at doubling steps, then combine.

StepOperationValue (exact to many digits)
12.0912.09
22.092 = 2.09 Γ— 2.094.3681
32.094 = (2.092)24.3681Β² = 19.08049961
42.098 = (2.094)219.08049961Β² = 363.936151... (approx)
5Combine to get 2.0910 = 2.098 Γ— 2.092363.936151... Γ— 4.3681 = 1590.2406878544866...

Final answer (a): (2.09)10 β‰ˆ 1590.2406878544866. Rounded to sensible significant figures: 1590.2407 or 1.5902407 Γ— 103.

Solution (b) β€” compute (0.5216)2/3

Interpretation: (0.5216)2/3 = βˆ›(0.5216Β²) or use logs: log y = (2/3) log(0.5216) then antilog.

StepOperationValue
1Square: 0.5216Β²0.5216 Γ— 0.5216 = 0.271265856
2Cube root: βˆ›(0.271265856)Numeric evaluation β‰ˆ 0.6479752060827903
Alternate (log) routeFind log10(0.5216), multiply by 2/3, take antilogSame numeric result β‰ˆ 0.6479752061

Final answer (b): (0.5216)2/3 β‰ˆ 0.6479752061 (β‰ˆ 0.64798 to 5 s.f.).

Real-life example: (a) Repeated-interest or compound growth model for 10 periods. (b) Fractional powers appear in physics: e.g., root-mean-square or scaling laws where you take fractional exponents of measured quantities.
Profit idea: Build an educational micro-service / mobile app that converts complex fractional powers and shows both the repeated-squaring table and log method; market to students and tutors with a small subscription.

Question 17

Task: By using mathematical tables evaluate:

  1. 40.5 Γ— 300 Γ— 0.008904 β€” (Note: this was Question 15 earlier; but the main Q17 in the list given later asks:)
In our original list Question 17 asks to evaluate:
  1. (8.802 Γ— 0.00123) Γ· (0.01252 Γ— 352080)
  2. √(2.16)3 Γ· (0.534 Γ— 3333) interpreted as (√(2.16))3 / (0.534 Γ— 3333) i.e. (2.16)3/2 / (0.534 Γ— 3333).

Symbols & method

  • Parentheses imply operations to be done first.
  • √(2.16)3 we treat as (√2.16)3 = (2.16)3/2.
  • When using logs/tables: compute logs of numerator and denominator separately, subtract, then antilog. Here we show direct numeric multiplication/division with digit-by-digit clarity.

Part (1)

StepWork
Numerator8.802 Γ— 0.00123 = multiply digit-by-digit:
8.802 Γ— 0.00123 = 8.802 Γ— (123 Γ— 10-5) = (8.802 Γ— 123) Γ— 10-5.
8.802 Γ— 123 = 8.802Γ—(100 + 20 + 3) = 880.2 + 176.04 + 26.406 = 1082.646 β†’ scale by 10-5 gives 0.01082646.
Denominator0.01252 Γ— 352080. Compute: 352080 Γ— 0.01252 = 352080 Γ— (1252 Γ— 10-5) = (352080 Γ— 1252) Γ— 10-5.
352080 Γ— 1252 = 352080Γ—(1000 + 200 + 50 + 2) = 352,080,000 + 70,416,000 + 17,604,000 + 704,160 = 4408,041, (check grouping) β†’ result = 4,408.04160 after scaling by 10-3 etc.
Direct decimal result: 4408.04160.
Division0.01082646 Γ· 4408.04160 β‰ˆ 0.0000024560702875399361.

Final answer 17(1): β‰ˆ 2.45607028754 Γ— 10-6 (decimal β‰ˆ 0.00000245607).

Part (2)

StepWork
Compute numerator(√2.16)3 = (2.16)3/2.
2.163/2 = (2.16)1.5 = 2.16 Γ— √2.16 β‰ˆ 2.16 Γ— 1.4702... β‰ˆ 3.1745387066469988.
Denominator0.534 Γ— 3333 = 0.534 Γ— (3333) = multiply: 3333Γ—0.5=1666.5; 3333Γ—0.03=99.99; 3333Γ—0.004=13.332 β†’ sum β‰ˆ 1779.822. Exact: 1779.822.
Division3.1745387066469988 Γ· 1779.822 β‰ˆ 0.0017836270743068682.

Final answer 17(2): β‰ˆ 1.7836270743 Γ— 10-3 (decimal β‰ˆ 0.00178362707).

Real-life example: These computations appear when normalising measurement units for instrument calibration β€” very small numerical ratios often determine accept/reject tolerances in quality control.
Profit idea: Create a precision-calculation plugin for manufacturing QC reports that produces table outputs like the above (automated traces + audit trail) and sell to local factories or labs.

Question 18

Task: Two concentric circles, area of ring A = Ο€ (R2 βˆ’ r2). Given R = 12.05 mm, r = 10.05 mm, Ο€ = 3.142. Use tables (or direct arithmetic).

Symbols

R = larger radius, r = smaller radius, Ο€ = 3.142 (given). Area of ring = difference of areas: Ο€RΒ² βˆ’ Ο€rΒ² = Ο€(RΒ² βˆ’ rΒ²).

StepWork
Compute RΒ²12.05 Γ— 12.05 β†’ digit-by-digit: 12.05 Γ— 12.05 = (12 + 0.05) Γ— (12 + 0.05) = 144 + 0.6 + 0.0025 = 144.6025.
Compute rΒ²10.05 Γ— 10.05 = (10 + 0.05)Β² = 100 + 1.0 + 0.0025 = 101.0025.
DifferenceRΒ² βˆ’ rΒ² = 144.6025 βˆ’ 101.0025 = 43.6000.
Multiply by Ο€A = 3.142 Γ— 43.6000. Multiply: 43.6 Γ— 3.142
43.6Γ—3 = 130.8; 43.6Γ—0.1 = 4.36; 43.6Γ—0.04 = 1.744; 43.6Γ—0.002 = 0.0872 β†’ sum = 130.8 + 4.36 + 1.744 + 0.0872 = 136.9912.

Final answer Q18: A = 136.9912 mmΒ² (β‰ˆ 136.991 mmΒ²).

Real-life example: Material needed for a circular washer (metal ring) where outer radius and inner radius are given β€” area tells you metal required per washer.
Profit idea: Offer a small fabrication estimator for local metal workshops that takes radii and material cost and returns material cost per washer β€” charge per quote or include in a paid app for small manufacturers.

Question 19

Task: Use mathematical tables to calculate T = 2Ο€ √(l/g), given l = 0.825, g = 9.81, and Ο€ = 3.142.

Symbols

T β€” period (seconds), l β€” length (m), g β€” acceleration due to gravity (m/sΒ²), Ο€ β€” 3.142. The formula is the standard small-angle pendulum period formula.

StepWork
Compute inside sqrtl/g = 0.825 Γ· 9.81:
0.825/9.81 β‰ˆ 0.084108846 (carry sufficient digits for accuracy).
Square root√(0.084108846) β‰ˆ 0.2900... (computed precisely β‰ˆ 0.2903805609).
Multiply by 2Ο€2 Γ— 3.142 Γ— 0.2903805609 β‰ˆ
2Γ—3.142 = 6.284. Then 6.284 Γ— 0.2903805609 β‰ˆ 1.8223368068.

Final answer Q19: T β‰ˆ 1.8223368068 s (β‰ˆ 1.82234 s, to 6 s.f.).

Real-life example: Designing a pendulum clock or timing experiment: a 0.825 m pendulum has a period β‰ˆ 1.82 s (time for one full oscillation).
Profit idea: Produce a simple 'physics experiment kit' for schools: include a pendulum, an instruction booklet with exact computations and worksheets; sell to schools and tutors.

Question 20

Task: Given 1/u + 1/v = 1/f, use logarithm tables to find u when v = 47.9 cm and f = 10.28 cm.

Symbols & meaning

  • u and v are object and image distances respectively (common in lens formula); f is focal length.
  • Equation rearranged for u:
    1/u = 1/f βˆ’ 1/v β†’ u = 1 / (1/f βˆ’ 1/v).

StepWork
Compute 1/f1 Γ· 10.28 β‰ˆ 0.0972677596 (digit-by-digit: 10.28 into 1 β†’ 0.0972...)
Compute 1/v1 Γ· 47.9 β‰ˆ 0.0208757860.
Subtract1/f βˆ’ 1/v = 0.0972677596 βˆ’ 0.0208757860 = 0.0763919736.
Reciprocalu = 1 Γ· 0.0763919736 β‰ˆ 13.0891015417 cm.

Final answer Q20: u β‰ˆ 13.0891015417 cm (β‰ˆ 13.089 cm to 4 decimal places).

Real-life example: In optics, if the image of an object forms 47.9 cm from a lens whose focal length is 10.28 cm, the object is β‰ˆ 13.09 cm from the lens β€” useful for lab setups and microscope focusing.
Profit idea: Build an app for small optics labs and hobbyists to input lens data and get object/image distances, ray diagrams, and parts lists β€” monetise via one-time purchase or pro features (printing, batch calculations).
Answers β€” Questions 21–25

Answers β€” Questions 21 to 25

Detailed steps, symbol explanations, small tables where helpful, real-life examples, and profitable ideas included.

Question 21

Task: Using the formula v2 = u2 + 2as, calculate s given u = 18.5, v = 36, a = 3.8.

Symbols & method

  • u β€” initial velocity (units: same as v).
  • v β€” final velocity.
  • a β€” acceleration (constant).
  • s β€” displacement (what we want).
  • Rearrange: s = (v2 βˆ’ u2) / (2a).

Step-by-step arithmetic (digit-by-digit)

StepComputation
1. Compute vΒ²v = 36 β†’ vΒ² = 36 Γ— 36 = 1296.
2. Compute uΒ²u = 18.5. Compute 18.5Β² carefully:
18.5 Γ— 18.5 = (18 + 0.5) Γ— (18 + 0.5) = 18Β² + 2Γ—18Γ—0.5 + 0.5Β² = 324 + 18 + 0.25 = 342.25.
3. SubtractvΒ² βˆ’ uΒ² = 1296 βˆ’ 342.25 = 953.75.
4. Compute denominator2a = 2 Γ— 3.8 = 7.6.
5. DivideCompute s = 953.75 Γ· 7.6. Multiply numerator & denominator by 10 for easier division: 9537.5 Γ· 76.
76 Γ— 125 = 9500 β†’ remainder 37.5. Then 37.5 Γ· 76 β‰ˆ 0.4934210526315789. So 9537.5 Γ· 76 = 125.49342105263158.

Final answer Q21: s β‰ˆ 125.4934211 (units same as velocities' distance unit). Rounded sensibly: 125.49.

Example: If a car speeds from 18.5 m/s to 36 m/s with constant acceleration 3.8 m/sΒ², it travels β‰ˆ 125.49 m during that acceleration.
Profit idea: Build a simple automotive diagnostics app for entry-level mechanics: input speed and acceleration, compute stopping/acceleration distances and recommend safe upgrade parts (sell as add-on guides).

Question 22

Task: Given V = Ο€ r2 h, find r when V = 64.91 cm3, h = 3.907 cm, and Ο€ = 3.142.

Symbols & rearrangement

  • V β€” volume of right circular cylinder.
  • r β€” radius of base (unknown).
  • h β€” height (given).
  • Rearrange: rΒ² = V / (Ο€ h), then r = √(V / (Ο€ h)).

Step-by-step arithmetic

StepWork
1. Compute Ο€ Γ— hΟ€ Γ— h = 3.142 Γ— 3.907. Do digit-by-digit:
3.142 Γ— 3.907 = 3.142Γ—(3 + 0.9 + 0.007) = 3.142Γ—3 (9.426) + 3.142Γ—0.9 (2.8278) + 3.142Γ—0.007 (0.021994) β†’ sum = 9.426 + 2.8278 + 0.021994 = 12.275794.
2. Compute rΒ² = V / (Ο€ h)rΒ² = 64.91 Γ· 12.275794.
Divide: 12.275794 Γ— 5 = 61.37897 (close). Remainder = 64.91 βˆ’ 61.37897 = 3.53103. Next fraction β‰ˆ 3.53103 / 12.275794 β‰ˆ 0.2877. So rΒ² β‰ˆ 5.2877.
3. Compute r = √(r²)Square root of 5.2877:
Note: 2.3Β² = 5.29 (very close). So r β‰ˆ 2.299 (more precisely β‰ˆ 2.2995).

Final answer Q22: r β‰ˆ 2.299 cm (rounded to 3 d.p.).

Example: Finding the radius of a cylinder container (e.g., a small jar) from its volume and height β€” useful when designing packaging to fit a specified volume.
Profit idea: Offer a product-design calculator for small manufacturers: enter desired volume/height and get radius, material area, and cost estimates; charge per quote or subscription.

Question 23

Task: Given (4/3)Ο€ r3 = 234.5, use tables to calculate 4Ο€ r2. (Use Ο€ = 3.142.)

Key observation & algebraic shortcut

Notice:

  • (4/3)Ο€ rΒ³ = 234.5 is the volume of a sphere formula rearranged.
  • If we compute 4Ο€ rΒ² directly, we can use the relation 4Ο€ rΒ³ = 3 Γ— 234.5 = 703.5 (multiply both sides of the given by 3).
  • Then 4Ο€ rΒ² = (4Ο€ rΒ³) / r = 703.5 / r. So we need r (cube root of the rΒ³ value). This is arithmetic + one cube root step.

Step-by-step arithmetic

StepWork
1. Compute 4Ο€ rΒ³From given: (4/3)Ο€ rΒ³ = 234.5. Multiply both sides by 3: 4Ο€ rΒ³ = 234.5 Γ— 3 = 703.5.
2. Solve for rΒ³rΒ³ = (703.5) / (4Ο€) = 703.5 Γ· (4 Γ— 3.142) = 703.5 Γ· 12.568. Compute 12.568 Γ— 56 = 703.808 (slightly too big), so rΒ³ β‰ˆ 703.5 Γ· 12.568 β‰ˆ 55.9755 (approx).
3. Cube root to find rFind r = βˆ›(55.9755). Estimate: 3.82Β³ = (3.82Β²)Γ—3.82 = 14.5924Γ—3.82 β‰ˆ 55.74 (a bit low). 3.826Β³ β‰ˆ 56.00 (a bit high). A good estimate is r β‰ˆ 3.8256.
4. Use 4Ο€ rΒ² = 703.5 / rNow divide:
4Ο€ rΒ² β‰ˆ 703.5 Γ· 3.8256 β‰ˆ 183.91 (rounded).

Final answer Q23: 4Ο€ rΒ² β‰ˆ 183.9 (same units squared as rΒ² Γ— constant). More precise β‰ˆ 183.91.

Example: If a sphere has volume 234.5 (unitsΒ³), the surface area (which is 4Ο€ rΒ²) is β‰ˆ 183.9 unitsΒ² β€” useful for estimating paint or coating required.
Profit idea: Create a simple tool for craft makers: input a mold's volume, get surface area and material costs for coating β€” market to small artisans and model makers.

Question 24

Task: Find x if (log2 x) (βˆ’3 + log2 x) = 4.

Symbols & strategy

  • Let y = log2 x. Then the equation becomes y(βˆ’3 + y) = 4.
  • Solve the quadratic in y, then convert back: x = 2y.

Step-by-step

StepWork
1. Substitutey(y βˆ’ 3) = 4 β†’ yΒ² βˆ’ 3y βˆ’ 4 = 0.
2. Factor quadraticyΒ² βˆ’ 3y βˆ’ 4 = (y βˆ’ 4)(y + 1) = 0. So y = 4 or y = βˆ’1.
3. Convert back to xIf y = log2 x then x = 2y:
  • For y = 4: x = 2⁴ = 16.
  • For y = βˆ’1: x = 2βˆ’1 = 1/2.

Final answer Q24: x = 16 or x = 1/2. (Both are valid because logarithm argument must be positive.)

Example: Logarithmic substitutions like y = log2 x let multiplicative/exponential relationships convert to algebraic quadratics β€” used when analyzing signal gains in dB or calculating scale factors.
Profit idea: Provide worked, stepwise online practice for high-school math exams β€” charge for bundles of fully solved problems with explanations and interactive checking.

Question 25

Task: Solve for x, given log9 x + 3 log3 3 = 4.

Symbols & simplification

  • log9 x is logarithm of x with base 9.
  • log3 3 is logarithm of 3 with base 3 β€” that equals 1 (since 3ΒΉ = 3).
  • So 3 log3 3 = 3 Γ— 1 = 3. The equation becomes log9 x + 3 = 4 β‡’ log9 x = 1.

Step-by-step

StepWork
1. Evaluate log3 3log3 3 = 1 because 3ΒΉ = 3.
2. SubstituteEquation becomes log9 x + 3 = 4 β†’ log9 x = 1.
3. Convert to exponential formlog9 x = 1 means 9ΒΉ = x β†’ x = 9.

Final answer Q25: x = 9.

Example: This type of manipulation is routine when changing bases in logarithms or simplifying expressions in financial models where different compounding bases appear (e.g., base 3 vs base 9 growth factors).
Profit idea: Create a short video/course focused on logarithm intuition (changing bases, practical examples) and sell it to tutors and students preparing for exams β€” include downloadable solved worksheets like these.

Reference Book: N/A

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