# 9th Class Math Chapter 3 Logarithms Short Questions Answer

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## 9th Class Math Chapter 3 Logarithms Short Questions Answer

Find the value of x in the following.

log₃ x = 5

3⁵ = x

x = 3⁵

x = 243

log₄ 256 = x

4x = 256

4x = 4⁴

x = 4

log₆₂₅ 5 = ¼ x

(625)1/4x = 5

(5⁴)1/4x = 5

5x = 5¹

x = 1

1. log₆₄ x = -2/3

(64)-2/3 = x

(4³)-2/3 = x

4⁻² = x

x = 1/4²

x = 1/16

Find the value of x in the following. Log x = 2.4543 Log x = 0.1821 Log x = 0.0044 Log x = 1.6238

Solution:

1. Log x = 2.4543

x = antilog 2.4543

From the table against the row of 0.45 under 4 we have 2844 and difference under 3 is 2. Adding 2844 and 0 we get 2846.

x = 284.6

1. Log x = 0.1821

x = antilog 0.1821

From the table against the row of 0.18 under 2 we have 1521 and difference under 1 is 0. Adding 1521 and 0 we get 1521.

x = 1.521

1. Log x = 0.0044

x = antilog 0.0044

From the table against the row of 0.00 under 4 we have 1009 and difference under 4 is 1. Adding 1009 and 1 we get 1010.

x = 1.010

1. Log x = 1.6238

x = antilog 1.6238

From the tale against the row of 0.62 under 3 we have 4198 and difference under 8 is 8. Adding 4192 and 8 we get 4206.

x = 0.04206

If log 2 = 0.3010, log 3 = 0.4771, and log5 = 0.6990, then find the value of the following.
Solution:

1. Log 45

= log (3 x 3 x 5)

= log 3 + log 3 + log 5

= 0.4771 + 0.4771 + 0.6990

= 1.6532

1. Log 16/15

= Log16 – log15

= Log2⁴ – log 3 x 5

= 4 log 2 – log 3 – log 5

= 4(0.3010) – 0.4771 – 0.6999

= 1.6532

1. Log 0.048

= log 48/1000

= log 48 – log1000

= log 3 x 16 – log 10³

= log 3 + log 16 – 3log 10

= 0.4771 + log 2⁴ – 3(1)

= 0.4771 + 4 log 2 – 3

= 0.4771 + 4 (0.3010) – 3

= 1.6811 – 3

= 1 + 0.6811 – 3

= – 2 + 0.6811

= 2.6811

Simplify the following.

Solution:

1. ³√25.47

Let x = ³√25.47

x = (25.47)1/3

Taking log on both sides

Log x = log (25.47)1/3

Log x = 1/3log 25.47

Log x = 1/3(1.4060)

Log x = 0.4686

Taking Antilog on both sides, we get

Antilog (Log x) = antilog(0.4686)

x = 2.942

1. ⁵√342.2

Let x = ⁵√342.2 = (342.2)1/5

Taking log on both sides

Log x = (342.2)1/5

Log x = 1/5 log 342.2

Log x = 1/5 (2.5343)

Log x = 0.5069

Taking Antilog on both sides, we get

Antilog (log x) = antilog (0.5069)

x = 3.213

1. (8.97)³ x (3.95)²/³√15.37

Let x = (8.97)³ x (3.95)²/³√15.37

x = (8.97)³ x (3.95)²/(15.37) 1/³

Taking log on both sides, we get

Log x = log(8.97)³ x (3.95)²/(15.37)1/³

Log x = log(8.97)³ x (3.95)² – log(15.37)1/³

Log x = log(8.97)³ + log(3.95)² – log(15.37)1/³

Log x = 3 log(8.97) + 2 log (3.95) – 1/3 log(15.37)

Log x = 3(0.9528) + 2(0.5966) – 1/3 (1.1867)

Log x = 2.8584 + 1.1932 – 0.3956

Log x = 3.6560

Taking antilog on both sides, we get

x = antilog (3.6560)

x = 4529

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