logarithm.docx

7/27/2019 Logarithm.docx

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PART 1

a) Logarithms were invented independently by John Napier, a Scotsman, and by Joost

Burgi, a Swiss. Napier's logarithms were published in 1614; Burgi's logarithms were

published in 1620.

The objective of both men was to simplify mathematical calculations. This approach

originally arose out of a desire to simplify multiplication and division to the level of addition

and subtraction. Of course, in this era of the cheap hand calculator, this is not necessary

anymore but it still serves as a useful way to introduce logarithms.

b)The first field of study: Chemistry

In chemistry, a solutions pH is defined by the logarithmic equation:

p(h) =

log10 (h)

where h is the hydronium ion concentration in moles per liter. We usually round pH values to

the nearest tenth.

Example:Find the pH of a solution with hydronium ion concentration 4.5 x 10-5Given h=4.5 x 10-5,

p(h) = log10 (4.5 x 10-5)= ( log10 4.5 + log10 10-5 )= ( log10 4.5 + (5)(log10 10) )= ( 0.6532 + (5) )= ( 0.6532 5 )= (4.3468)= 4.3


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The second field of study: Physics

The loudness of sound is measured in units called decibels. These units are measured by

first assigning an intensity I0 to a very soft sound (which is called the threshold sound). The

sound we wish to measure is assigned an intensity I, and we measure the decibel rating d of

this sound with the equation.

Example:

Find the decibel rating of a sound with intensity 5000I0.

d = 10 log10

= 10 log10 5000

= 36.99 decibels


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PART 2

(a) i) By using a pair of vernier calipers, the diameter of six (6) different

spheres is taken.

ii) The volume of spheres is obtained by using water displacement method.

iii) The values of the diameter, D, in cm, and its corresponding volume, V, cm3

are tabulated as below:

Diameter (cm) Volume (cm )

2 4.19

3 14.14

4 33.52

5 65.48

6 113.14

7 179.67


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b) When diameter, d = 4cm, volume, v = 33.52 cm333.52 = m x 4n

log10 33.52 = log10 (m x 4n)

log10 33.52 = log10 m + log10 4n

log10 33.52 = log10 m + n log10 4

1.525 = log10 m + 0.602n

When diameter, d = 7cm, volume, v = 179.67 cm3179.67 = m x 7n

log10 179.67 = log10 (m x 7n)

log10 179.67 = log10 m + log10 7n

log10 179.67 = log10 m + n log10 7

2.254 = log10 m + 0.845n

By forming the simultaneous equation,

1.525 = log10 m + 0.602n2.254 = log10 m + 0.845n

0.729 = 0.243n

n = 0.729 / 0.243

n = 3

By substituting n= 3 into one of the equation,

1.525 = log10 m + 0.602 x 3

log10 m = 1.525 1.806

log10 m = 0.281

m = 0.524


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PART 3

(A) Based on the table in PART 2 (a) (ii), graph V against D is drawn.

Diameter (cm) Volume (cm )

2 4.193 14.14

4 33.52

5 65.48

6 113.14

7 179.67

-

20.00

40.00

60.00

80.00

100.00

120.00

140.00

160.00

180.00

200.00

0 1 2 3 4 5 6 7 8

V AGAINST D


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(B) a)To reduce the equation V = mDn to a linear form,

b)

(c) i) Based on the graph in (B) (b),

Gradient, m =

= 3

n = m

n = 3


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y-intercept, c =0.28

log10 m = c

log10 m =0.28

m = 0.525

ii) When the diameter, D = 5 cm, log10 5 = 0.699

Based on the graph in (B) (b),

log10 V = 1.816

V = 65.48 cm3

iii) When the volume, V = 180 cm3, log10 180 = 2.255

Based on the graph in (B) (b),

log10 D = 0.85

D = 7.08 cm


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logarithm.docx

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