math class 9 new book chapter 2 mcqs | Notes Punjab
math class 9 new book chapter 2 mcqs

MCQs 9th math ch2 | 2025

MCQs 9th math ch2 2025

MCQs 9th math ch2 2025 will be solved in complete detail. Name of chapter 2 is logarithms. New book math class 9 chapter 2 is very important with respect to MCQs.

Q 1 Four option are given against each statement. Encircle the correct option.

(i) $$The\;s\tan dard\;form\;of\\5.2\;\times\;10^6\;is:$$
(a) 52,000 (b) 520,000 (c) 5.200,000 (d) 52,000,000

Option C is correct.

——————————————————————————-

(ii) Scientific notation of 0.00034 is :
$$(a)\;3.4\times10^3\;(b)\;3.4\times10^{-4}\\(c)\;3.4\times10^4\;(d)\;3.4\times10^{-3}$$

Option B is correct.

——————————————————————————-

(iii) The base of common logarithm is:
(a) 2  (b) 10  (c) 5  (d) e

Option B is correct.

Explanation:- Base of common logarithm is 10 and base of natural logarithm is e.

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(iv)  $$Log_2\;2^3\;=\;$$
(a) 1  (b) 2  (c) 5  (d) 3

Option D is correct.

Explanation:- First we will apply power rule of logarithms.

$$Log_2\;2^3\;=\;3\;Log_2\;2$$

Log of any number with the same base will give us 1

= 3 ( 1 ) = 3

——————————————————————————-

(v)  Log 100 =
(a) 2  (b) 3  (c) 10  (d) 1

Option A is correct.

Explanation:- Log 100 = Log ( (10) (10 )

Apply  Log m n  =  Log m  + Log n

= Log 10 + Log 10

= 1 + 1 + 2

——————————————————————————-

(vi) If  Log 2 = 0.3010, then  Log 200 is:
(a) 1.3010  (b) 0.6010  (c) 2.3010  (d) 2.6010

Option C is correct.

Explanation:- Log 200 = Log ( (2) (100) )

After applying product rule of logarithms, we get

= Log 2 + Log 100

= 0.3010 + 2 = 2.3010

——————————————————————————-

(vii)  Log (0) =
(a)  Positive  (b)  Negative  (c) Zero  (d)  Undefined

Option D is correct.

Explanation:- Any number raised to any power will not give us zero.

$$Log_{10}0\;=\;x\\\\means\;10^x\;=\;0$$

There is no such x which gives 0.


(viii) Log 10,000 =
(a) 2  (b) 3  (c) 4  (d) 5

Option C is correct.

Explanation:-

$$Log\;10000\;=\;Log\;10^4$$

Lets use power rule of logarithms

= 4 Log 10 = 4 (1) = 4

——————————————————————————-

(ix)  Log 5 + Log 3 =
(a) Log 0   (b) Log 2   (c) Log (5/3)   (d) Log 15

Option D is correct.

Explanation:-

We will solve it with the help of product law of logarithms.

Log m n = Log m + Log n

Log 15 = Log ( (5) (3) ) = Log 5 + Log 3

——————————————————————————-

(x)  $$3^4\;=\;81\;in\;\log arithmic\;form\;is:$$
$$(a)\;Log_34\;=\;81\;\;(b)\;Log_43\;=81\\(c)\;Log_381\;=4\;\;(d)\;Log_481=\;3$$

Option C is correct.

Explanation:- 

Base will not be changed either we want to convert logarithmic form into exponential form or exponential form into logarithmic form.

$$Exponential\;form\\a^x\;=\;y\\\\Logarithmic\;form\\Log_ay\;=\;x$$

As you can see above that the base is ‘a’ both in exponential and logarithmic form.

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