Concept and Application of Logarithm (Class- 9, Experience- 3)

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• Exercise Solution

[Exercise- page 80]

E.3.1. Find out the value using different formulas. 

i)  $\mathbf{2}\sqrt[\mathbf{3}]{\mathbf{343}}\mathbf{+}\mathbf{2}\sqrt[\mathbf{5}]{\mathbf{243}}\mathbf{-}\mathbf{12}\sqrt[\mathbf{6}]{\mathbf{64}}$ 

ii)  y^{a + b}  y^2c  ✕  y^{b + c}  y^2a  ✕  y^{c + a}  y^2b

Solution:

i)  $\mathbf{2}\sqrt[\mathbf{3}]{\mathbf{343}}\mathbf{+}\mathbf{2}\sqrt[\mathbf{5}]{\mathbf{243}}\mathbf{-}\mathbf{12}\sqrt[\mathbf{6}]{\mathbf{64}}$ 

= 2 ³√(7)³ + 2 ⁵√(3)⁵ – 12 ⁶√(2)⁶

= 2(7³)¹³ + 2(3⁵)¹⁵ – 12(2⁶)¹⁶

= 2 ✕ 7 + 2 ✕ 3 – 12 ✕ 2 

= 14 + 6 – 24 

= -4 (Answer)

ii)  y^{a + b}  y^2c  ✕  y^{b + c}  y^2a  ✕  y^{c + a}  y^2b

= y^a+b-2c ✕ y^b+c-2a ✕ y^c+a-2b

= y^{a+b-2c+b+c-2a+c+a-2b}

= y⁰

= 1 (Answer)

E.3.2. Using different formulas, prove, 

(z^az^b )^{a + b - c}  ✕  (z^bz^c )^{b + c - a}  ✕  (z^cz^a )^{c + a - b}

Solution:

(z^az^b )^{a + b - c}  ✕  (z^bz^c )^{b + c - a}  ✕  (z^cz^a )^{c + a - b}

= z^(a-b)(a+b-c) ✕ z^(b-c)(b+c-a) ✕ z^(c-a)(c+a-b)

= z^{(a-b)(a+b-c) + (b-c)(b+c-a) + (c-a)(c+a-b)}

Here, 

(a-b)(a+b-c) + (b-c)(b+c-a) + (c-a)(c+a-b) 

= (a² - ab + ab - b² - ca + bc) + (b² - bc + bc - c² - ab + ca) + (c² - ca + ca - a² - bc + ab)

= (a² - b² - ca + bc) + (b² - c² - ab + ca) + (c² - a² -bc + ab) 

= (a² - b² + b² - c² + c² - a²) + (-ca + bc – ab + ca – bc + ab) 

= 0 + 0 

= 0

So, 

    z^{(a-b)(a+b-c) + (b-c)(b+c-a) + (c-a)(c+a-b)} 

 = z⁰ 

 = 1     [proved]

E.3.3. Express the following exponential equalities in log and find out the value of x using a digital device. 

i) 2^x = 64

ii) (1.2)^x = 100

iii) 7^x = 5

iv) (23)^x = 7

Solution:

(i) 2^x = 64 

     2^x = 64 

or, log₂(2^x) = log₂(64) [In both side taken log₂] 

or, log₂ (2^x) = log₂(64) 

or, x.log₂2 = log₂(64) 

or, x.1 = log₂(64) [Rule is, log_aa = 1] 

or, x.1 = 6 [By using scientific calculator] 

or, x = 6   [Answer] 

(ii) (1.2)^x = 100 

     (1.2)^x = 100 

or, log_1.2(1.2^x) = log_1.2(100) [In both side taken log_1.2] 

or, x.log_1.21.2 = log_1.2(100) 

or, x.1 = log_1.2(100) [Rule is, log_aa = 1] 

or, x.1 = 25.2585 (Approx.) [By using scientific calculator] 

or, x = 25.2585 (Approx.)   [Answer] 

(iii) 7^x = 5 

     7^x = 5 

or, log₇(7^x) = log₇(5) [In both side taken log₇] 

or, log₇(7^x) = log₇(5) 

or, x.log₇7 = log₇(5)

or, x.1 = log₇(5) [Rule is, log_aa = 1]

or, x.1 = 0.8271 [By using scientific calculator]

or, x = 0.8271 (Approximately)  [Answer]

(iv) (23)^x = 7

     (23)^x = 7

or, log_2/3(23)^x = log_2/3(7) [In both side taken log_2/3]

or, x.log_2/3(23) = log_2/3(7)

or, x.log_2/3(23) = log_2/3(7)

or, x.1 = log_2/3(7) [Rule is, log_aa = 1]

or, x.1 = -4.799  [Find value by using scientific calculator]

or, x = -4.799 (Approximately) [Answer]

E.3.4. In how many years will the compound capital be 3 times at 10% compound interest rate?

Solution:

Lets say, 

     Initial capital = P, 

     Compound capital, A = 3P and 

     Compound Interest Rate, r = 10% =  10100 = 0.1

According to Formula, 

3P = P(1 + 0.1)ⁿ   [Compound Formula, A=P(1+r)ⁿ ] 

or, 3 = (1+0.1)ⁿ   [both side divided by P]

or, 3 = (1.1)ⁿ 

or, log_1.13 = log_1.1(1.1)ⁿ

or, n. log_1.1(1.1) = log_1.13    [Side change]

or, n. 1 = log_1.13   [we know, log_aa = 1]

or, n ≈ 11.5267    

So, Initial capital will be triple by approximately 11.5267 years.  (Answer)

E.3.5. You all know the name of corona virus. This virus spreads quickly. If corona virus spreads from 1 person to 3 persons per day, how many people will be infected with corona virus in 1 month? After how many days, 1 crore People will be infected?

Solution:

To solve this problem, we can use exponential growth since the number of infected people triples each day.

Let's start with 1 person infected.

Day 1: 1 person = 3⁰

Day 2: 3 people (1 ✕ 3) = 3¹

Day 3: 9 people (3 ✕ 3) = 3²

Day 4: 27 people (9 ✕ 3) = 3³

...

So, the number of infected people follows the pattern of 3ⁿ, where n is the number of days.

Now, we want to find out how many people will be infected in 1 month, which is approximately 30 days.

Let's calculate for day 30:

Number of infected people = 3³⁰ ≈ 205891132

so, number of infected people in 30 days is 205891132 (Answer)

Now, let's find out 2nd part of question, after how many days 1 crore (10000000) people will be infected.

We need to solve the equation:

3^x = 10000000

Or, log(3) ^x = log(10000000) [Taking the logarithm of both sides]

Or, x . log(3) = log(10000000)

Or, x = log(10000000)     log(3)

Or, x =       70.47121   [Use a scientific calculator]

Or, x ≈ 14.52 days

So, approximately after 14.52 days, 1 crore people will be infected with the coronavirus. (Answer)

E.3.6. Setu’s uncle has 3 bighas of land. He applies 30 kg of organic fertilizer every year to maintain the fertility of his land. If each kg of fertilizer increases the fertility of the land by 3%, find the depreciation of the land of Setu’s uncle? If he does not apply fertilizer to the land, then after how many years will his land have no crops?

Solution:

Answer will be posted soon. Please wait...

E.3.7. On 8 July 1918, an earthquake with a magnitude of 7.6 took place at Srimangal in Moulvibazar. Another earthquake with a magnitude of 6.0 took place in Chattogram on 22 November 1997. How many times stronger was the Srimangal earthquake than the Chattogram earthquake?

Solution:

We can use logarithms to solve this problem. Since the Richter scale is logarithmic, we can use the relationship between the magnitude difference and the difference in energy released.

Given:

Earthquake of Srimangal has a magnitude, I₁ = 7.6

Earthquake of Chattogram has a magnitude, I₂ = 6.0.

The formula relating the difference in magnitudes to the ratio of the energy released is:

Difference in strength=10^ΔM [Where ΔM is the difference in magnitudes]

So, to find the difference in strength:

ΔM = Magnitude of Srimangal − Magnitude of Chattogram

ΔM = I₁ − I₂

ΔM = 7.6 − 6.0

ΔM = 1.6

Now, we plug this value into the formula:

Difference in strength = 10^1.6

Using logarithmic properties, we can calculate this value:

Difference in strength ≈ antilog(1.6) [To find value use calculator]

Difference in strength ≈ 39.81

So, earthquake of Srimangal is approximately 39.81 times stronger than earthquake of Chattogram. (Answer)

E.3.8. Once, an earthquake with a magnitude of 8 took place in Japan. In the same year, another earthquake took place which was 6 times stronger than the previous one. What was the magnitude of the later earthquake on the Richter scale?

Solution:

The Richter scale is logarithmic, meaning that each whole number increase represents a tenfold increase in amplitude of the seismic waves.

If the first earthquake in Japan had a magnitude of 8, and the second one in Japan was 6 times stronger, we can calculate its magnitude as follows:

Magnitude of the second earthquake = Magnitude of the first earthquake + log10(6)

Magnitude of the second earthquake = 8 + log10(6)

Using a calculator:

Magnitude of the second earthquake ≈ 8 + 0.778 = 8.778

So, the magnitude of the later earthquake on the Richter scale would be approximately 8.778. (Answer)

E.3.9. In July 1999, an earthquake with a magnitude of 5.2 took place at Maheshkhali in Cox’s Bazar. On 6 February 2023, an earthquake took place in the southern part of Turkey which was 398 times stronger than the Maheshkhali earthquake. What was the magnitude of the Turkey earthquake?

Solution:

To find the magnitude of the later earthquake on the Richter scale, we can use the fact that the Richter scale is logarithmic, where each whole number increase represents a tenfold increase in amplitude of the seismic waves.

If Coxbazer's earthquake had a magnitude of 5.2, and Turkey's one was 398 times stronger, we can calculate its magnitude as follows:

Magnitude of the Turkey's earthquake = Magnitude of Coxbazer's earthquake + log10(398)

Magnitude of the Turkey's earthquake = 5.2 + log10(398)

Using a calculator:

Magnitude of the Turkey's earthquake ≈ 5.2 + 2.6 = 7.8

So, the magnitude of the Turkey's earthquake on the Richter scale would be approximately 7.8 (Answer)