Review Exercise 2 2025 Math class 9
Review Exercise 2 2025 Math class 9 will be completely solved. Lets first solve the MCQs of class 9 math Review Exercise 2 new book. New book math class 9 chapter 2 review exercise 2.
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
Correct option is C.
(ii) Scientific notation of 0.00034 is :
$$(a)\;3.4\times10^3\;(b)\;3.4\times10^{-4}\\(c)\;3.4\times10^4\;(d)\;3.4\times10^{-3}$$
Correct option is B.
(iii) The base of common logarithm is:
(a) 2 (b) 10 (c) 5 (d) e
Correct option is B.
(iv) $$Log_2\;2^3\;=\;$$
(a) 1 (b) 2 (c) 5 (d) 3
Correct option is D.
(v) Log 100 =
(a) 2 (b) 3 (c) 10 (d) 1
Correct option is A.
(vi) If Log 2 = 0.3010, then Log 200 is:
(a) 1.3010 (b) 0.6010 (c) 2.3010 (d) 2.6010
Correct option is C.
(vii) Log (0) =
(a) Positive (b) Negative (c) Zero (d) Undefined
Correct option is D.
(viii) Log 10,000 =
(a) 2 (b) 3 (c) 4 (d) 5
Correct option is C.
(ix) Log 5 + Log 3 =
(a) Log 0 (b) Log 2 (c) Log (5/3) (d) Log 15
Correct option is D.
(x) $$3^4\;=\;81\;in\;\log arithmic\;form\;is:$$
$$(a)\;Log_34\;=\;81\;\;(b)\;Log_43\;=81\\(c)\;Log_381\;=4\;\;(d)\;Log_481=\;3$$
Correct option is C.
Q2 Express the following numbers in scientific notation:
$$(i)\;0.000567\;\;(ii)\;734\\(iii)\;0.33\;\times10^3$$
$$0.000567\;=5.67\;\times\;10^{-4}\;$$
$$734\;=\;7.34\;\times\;10^2$$
$$0.33\;\times\;10^3\;=\;3.3\times10^{-1}\times10^3\\=3.3\times10^{-1+2}\;=\;3.3\;\times\;10^2$$
Q3 Express the following numbers in ordinary notation:
$$(i)\;2.6\;\times\;10^3\;(ii)\;8.794\;\times\;10^{-4}\\(i)\;6\;\times\;10^{-6}$$
$$(i)\;2.6\;\times\;10^3\;=\;2600\\\\(ii)\;8.794\;\times\;10^{-4}=0.0008794\\\\(i)\;6\;\times\;10^{-6}=\;0.000006$$
Q4 Express each of the following in logarithmic form:
$$(i)\;3^7=2187\;(ii)\;a^b=c\\(iii)\;{(12)}^2\;=\;144$$
$$(i)\;3^7=2187\;\\Logarithmic\;form\;\;Log_32187=7\\\\(ii)\;a^b=c\\Logarithmic\;form\;\;Log_ac=b\\\\(iii)\;{(12)}^2\;=\;144\\Logarithmic\;form\;\;Log_{12}144=2$$
Q5 Express each of the following in exponential form:
$$(i)\;\log_48=x\;(ii)\;\log_9729=3\\(iii)\;\log_41024=5$$
$$(i)\;\;\log_48=x\;\\Exponential\;form\;\;4^x=8\\\\(ii)\;\;\log_9729=3\\Exponential\;form\;\;9^3=729\\\\(iii)\;\;\log_41024=5\\Exponential\;form\;\;4^5=1024$$
Q6 Find the value of x in the following:
$$(i)\;\;Log_9\;x=\;0.5$$
$$\;\;Log_9\;x=\;0.5\\Exponential\;form\;\;9^{0.5}\;=\;x\\\sqrt9\;=\;x\\x\;=\;3$$
$$(ii)\;\;{(\frac19)}^{3x}\;=\;27$$
$$\;\;{(\frac19)}^{3x}\;=\;27\\\;\;{(\frac1{3^2})}^{3x}\;=\;{(3)}^3$$
$${(3^{-2})}^{3x}\;=\;{(3)}^3$$
Law of power of power of indices will be used.
$${(3)}^{-2\times3x}\;=\;{(3)}^3\;0r\;{{(3})}^{-6x}\;=\;{(3)}^3$$
In equality bases are same, powers will also be same.
-6x = 3
-6/3 = x
x = -2
$$(iii)\;\;{(\frac1{32})}^{2x}\;=\;64$$
$$\;\;{(\frac1{32})}^{2x}\;=\;64$$
$${(\frac1{2^5})}^{2x}\;=\;{(2)}^6$$
$${(2^{-5})}^{2x}\;=\;{(2)}^6$$
Law of indices tells us that if a single number has two powers, powers will be multiplied.
$${(2)}^{-5\times2x}\;=\;{(2)}^6$$
$${(2)}^{-10x}\;=\;{(2)}^6$$
In equality, same bases implies same powers.
-10x = 6
-10/6 = x
x = -5/3
Q7 Write the following as a single logarithms:
$$(i)\;\;7\;Log\;x\;-\;3\;Log\;y^2$$
Law of power of logarithms will be used.
$$=\;Log\;x^7\;-\;Log\;{\;(y^2})^3$$
Let’s multiply the exponents ( powers ) i.e use law of indices.
$$=\;Log\;x^7\;-\;Log\;{\;y^6}$$
Quotient law of logarithms will be used.
$$=\;Log\;(\;\frac{x^7\;}{y^6})$$
(ii) 3Log4 – Log 32
Lets use Power rule of logarithms.
$$=\;Log\;4^3\;-\;Log\;32$$
= Log 64 – Log 32
Let’s use Quotient rule of logarithms.
= Log ( 64/32 ) = Log 2
$$(iii)\;\frac13(Log_58+Log_527)-Log_53$$
$$=\;\;\frac13Log_52^3+\frac13Log_53^3-Log_53$$
Now, we will use rule of power of indices.
$$=\;\frac13(3)Log_52+\frac13(3)Log_53-Log_53$$
$$=\;Log_52\;+\;Log_53\;-\;Log_53$$
$$=\;Log_52$$
Q8 Expand the following using laws of logarithms:
$$(i)\;\;Log\;(\;x\;y\;z^6\;)$$
Lets use product rule of logarithms, first.
$$=\;Log\;x\;+\;Log\;y\;+\;Loz\;z^6$$
Now we will use power rule of logarithms.
= Log x + Log y + 6 Log z
$$(ii)\;\;\log_3\sqrt[6]{m^5n^3}$$
First, we will use power rule of logarithms.
$$=\;\frac16\;\log_3\;m^5n^3$$
Product rule of logarithms. Log m n = Log m + Log n
$$=\;\frac16\;\lbrack\;Log_3\;m^5\;+\;Log_3\;n^3\;\rbrack$$
Again, we will use power rule of logarithms.
$$=\;\frac16\;\lbrack\;5\;Log_3\;m\;+3\;Log_3\;n\;\rbrack$$
$$=\;\frac56\;Log_3\;m\;+\frac12\;Log_3\;n$$
$$(iii)\;\;Log\;\sqrt{8\;x^3}$$
Square root means 1/2. Let’s use power rule of logarithms.
$$=\;\;\frac12\;Log\;8\;x^3$$
Let’s use Log m n = Log m + Log n
$$=\;\;\frac12\lbrack\;Log\;8\;+\;Log\;x^3\;\rbrack$$
$$=\;\;\frac12\lbrack\;Log\;2^3\;+\;Log\;x^3\;\rbrack$$
Again, let’s use power rule of logarithms.
$$=\;\;\frac12\lbrack\;3\;Log\;2\;+\;3\;Log\;x\;\rbrack$$
$$=\;\;\frac32\lbrack\;Log\;2\;+\;Log\;x\;\rbrack$$
Q9 Find the values of the following with the help of logarithm table:
$$(i)\;\;\sqrt[3]{68.24}$$
$$Let\;\;x=\;\sqrt[3]{68.24}$$
Taking common log on both the sides
$$Log\;x=Log\;\;\sqrt[3]{68.24}$$
Third root means power 1/3. Let’s apply power rule of logarithms.
$$Log\;x=\frac13\;Log\;68.24$$
Characteristic is 1 because there are two digits before decimal.
For mantissa we will check log table and in the log table mantissa is 8338 + 3 = 8341
$$Log\;x=\frac13\;(\;1.8341\;)$$
Log x = 0.6113
Taking antilog on both the side
x = Antilog ( 0.6113 )
x = 4.086
(ii) ( 319.8 ) ( 3.543 )
Let x = ( 319.8 ) ( 3.543 ) )
Taking common log on both the sides.
Log x = Log ( ( 319.8 ) ( 3.543 ) )
We will apply product rule of logarithms.
Log m n = Log m + Log n
Log x = Log 319.8 + Log 3.543
Characteristic of 319.8 is 2 because there are three digits before the decimal.
For Mantissa we will ignore the decimal.
In the log table mantissa of 3198 is 5038 + 11 = 5049
Characteristic of 3.543 is 0 because there is only one digit before the decimal and in the log table mantissa of 3543 is 5490 + 4 = 5494
Log x = 2.5049 + 0.5494
Log x = 3.0543
Taking antilog on both the sides.
x = Antilog 3.0543
x= 1133
$$(iii)\;\;\frac{36.12\;\times750.9}{113.2\;\times\;9.98}$$
Let $$x\;=\;\;\frac{36.12\;\times750.9}{113.2\;\times\;9.98}$$
Taking log on both the sides
$$Log\;x\;=\;Log\;\frac{36.12\;\times750.9}{113.2\;\times\;9.98}$$
Now, we will apply product and quotient rule of logarithms.
Log x=Log36.12+Log750.9-Log113.2-Log9.98
Characteristic of 36.12 is 1 because there are two digits before the decimal and characteristic of 750.9 is 2 because there are three digits before the decimal. Characteristic of 113.2 is 2 because there are three digits before the decimal and characteristic of 9.98 is 0 because there is only one digit before the decimal.
Mantissa of 36.12 is 5575 +2 = 5577 and mantissa of 750.9 is 8751 + 5 = 8756. Mantissa of 113.2 is 0531 + 8 = 0539 and mantissa of 9.98 is 9991.
Log x = 1.5577 + 2.8756 – 2.0539 – 0.9991
Log x = 1.3803
Taking antilog on both the sides.
x = Antilog 1.3803
x = 24.01
If you want to check how to find characteristics and mantissa from the log table, check on the link of the following video
How to find log with the help of log table
Q 10 In year 2016, the population of a city was 22 millions and was growing at a rate of 2.5 % per year. The function P(t) = 22 multiplied by 1.025 raised to power t gives the population in millions, t years after 2016. Use the model to determine in which year the population will reach 35 millions. Round the answer nearest year.
$$p\;(t)\;=\;22\;{(\;1.025\;)}^t$$
After putting the values of P(t) , we get
$$35\;=\;22\;{(\;1.025\;)}^t$$
Taking common log on both the sides
$$Log\;35\;=\;Log\;\{22\;{(\;1.025\;)}^t\}$$
Applying Log m n = Log m + Log n
$$Log\;35\;=\;Log\;22\;+\;Log\;{(\;1.025\;)}^t$$
Let’s apply power rule of logarithms.
$$Log\;35\;-\;Log\;22\;=\;t\;Log\;1.025$$
$$\frac{Log\;35\;-\;Log\;22}{\;Log\;1.025}\;=\;t$$
$$t\;=\;\frac{1.5441\;-\;1.3424}{0.0107}$$
After simplification, we get
t = 18.85
Hence after 19 years the population will reach 35 millions.
