9th Math Exercise 2.4 New book
9th Math Exercise 2.4 from new book is the most important exercise of chapter 2 Logarithms. Lets instantly check the solution 9th Math Exercise 2.4 new book Punjab boards.
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Q1 Without using calculator evaluate the following:
(i) $$\underset2{Log\;18}\;-\;\underset2{Log\;9}$$
$$\underset2{=\;Log\;(9\times2)}\;-\;\underset2{Log\;9}$$
Rule: Log (m n) = Log m + Log n
$$=\;Log_29\;+\;Log_22\;-\;Log_29$$
$$=\;Log_22\;=\;1$$
Log of any number with same base will give us 1.
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(ii) $$\underset2{Log\;64}\;+\;\underset2{Log\;2}$$
We know that Log of any number with same base will give us 1.
$$=\;Log_2\;{(2)}^6\;+\;1$$
$$Rule:\;\;Log\;m^n\;=\;n\;Log\;m$$
$$=\;6\;\underset2{Log}\;2\;+\;1$$
Log of any number with same base will give us 1.
=6 (1 ) +1 = 6+1 = 7
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(iii) $$\frac13\underset3{Log}8-\underset3{Log}18$$
$$=\;\frac13\;Log_3{(2)}^3\;-\;Log_3(2\times3\times3)$$
We will use power rule and product rule of Logarithm.
$$=\;\frac13(3)\;Log_32\;-\;\{\;Log_32\;+\;Log_33\;+\;Log_33\;\}$$
Log of a number with the same base will give us 1.
$$=\;Log_32\;\;-\;\{\;Log_32\;+\;1+\;1\;\}$$
$$=\;Log_32\;-\;Log_32\;-\;2\;=\;-2$$
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(iv) 2 Log 2 + Log 25
$$=\;2\;Log\;2\;+\;Log\;5^2\\\\=\;2\;Log\;2\;+\;2\;Log\;5\\\\=\;2\;(\;Log\;2\;+\;Log\;5\;)\\\\=\;2\;Log\;(\;2\times5\;)\\\\=\;2\;Log\;10$$
No Base is given, Hence base will be 10 and log of any number with the same base will give us 1.
$$=\;2\;Log_{10}10\\\\=\;2\;(\;1\;)\;=\;2$$
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(v) $$\;\frac13\;\log_4\;64\;+\;2\;Log_5\;25$$
$$=\frac13\;\log_4\;4^3\;+\;2\;Log_5\;5^2\\\\=\;\frac13(3)\;\log_4\;4\;+\;2\;(2)\;Log_5\;5\\\\=\;Log_4\;4\;+\;4\;Log_5\;5$$
Log of a number with the same base results 1.Hence,
= 1 + 4(1) = 1+4 = 5
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(vi) $$Log_3\;12\;+\;Log_3\;0.25$$
$$=\;Log_3\;(2^2\;\times\;3\;)\;+\;Log_3\frac{25}{100}$$
Rule: Log m n = Log m + Log n
$$=\;Log_3\;2^2\;+\;Log_3\;3\;+\;Log_3\frac14$$
we will use power and division rule of log and log of a number with same base gives 1.
$$=\;2\;Log_3\;2\;+\;1\;+\;Log_3\;1\;-\;Log_3\;4$$
Log of 1 with any base results zero because any number raised to power zero results 1.
$$=\;2\;Log_3\;2\;+\;1\;+\;0\;-\;Log_3\;2^2$$
Now we will use power rule of logarithm.
$$=\;2\;Log_3\;2\;+1\;-\;2\;Log_3\;2\\=\;1\\$$
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Q2 Write the following as a single Logarithm:
(i) (1/2) Log 25 + 2 Log 3
First, we will use power rule of logarithm.
$$=\;\frac12\;Log\;5^2\;+\;\;Log\;3^2\\\\\;=\;\frac12(2)\;Log\;5\;+\;\;Log\;9\\\\$$
= Log 5 + Log 9
We will us now Log m + Log n = Log m n
= Log ((9)(5))
= Log 45
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(ii) Log 9 – Log (1/3)
We will use quotient rule of Logarithm.
= Log 9 – ( Log 1 – Log 3 )
= Log 9 – Log 1 + Log 3
We know that Log 1 with any base results 0.
= Log 9 + 0 + Log 3
Now, we will use product rule of Logarithm.
= Log ((9) (3)) = Log 27
For different methods you can check links of Pdf and also You tube lectures given at the end of the lecture.
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(iii) $$Log_5\;b^2.\;Log_a\;5^3\\\\$$
Here, we cannot use product rule of logarithm i.e
Log m n = Log m + Log n
We will use power rule of logarithm.
$$=\;2Log_5\;b.\;3Log_a\;5\\\\=\;6\;Log_5\;b.\;Log_a\;5\\\\$$
Now, we will use change of base rule of logarithm.
$$Change\;of\;Base\;Rule\\\\Log_b\;a\;=\;\frac{Log_c\;a}{Log_c\;b}\\\\$$
$$=\;6\;\;\frac{Log\;b}{Log\;5}.\;\frac{Log\;5}{Log\;a}\\\\=\;6\;\;\frac{Log\;b}{Log\;a}\\\\=\;6\;\;Log_a\;b$$
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(iv) $$2\;Log_3x\;+\;Log_3y$$
We will use power and product rule of logarithm.
$$=\;Log_3x^2\;+\;Log_3y\\\\=Log_3\;x^2y$$
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(v) $$4\;Log_5x\;-\;Log_5y\;+\;Log_5z$$
$$=\;Log_5x^4\;-\;Log_5y\;+\;Log_5z\\\\=\;Log_5(\frac{x^4z}y)$$
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(vi) 2lna + 3lnb -4lnc
$$=\;\ln a^2\;+\;\ln b^3\;-\;\ln c^4\\\\=\ln(\frac{a^2\;b^3}{c^4})$$
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Q3 Expand the following using laws of logarithms:
(i) Log (11/5)
= Log 11 – Log 5
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(ii) $$Log_5\sqrt{8a^6}$$
First we will use power rule and then product rule of logarithms.
$$=\;\frac12\;\log_5\;8a^6\\=\;\frac12(\;Log_5\;8\;+\;Log_5\;a^6\;)$$
$$=\;\frac12(\;Log_5\;2^3\;+\;\;Log_5\;a^6\;)$$
Again we will use Power rule of Logarithms.
$$=\;\frac12\;(\;3\;Log_5\;2\;+\;6\;Log_5\;a\;)\\\\=\;\frac12\;(\;3\;Log_5\;2\;)+\;\frac12\;(\;6\;Log_5\;a\;)\\\\=\;\frac32\;Log_5\;2\;+\;3\;Log_5\;a\\$$
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(iii) $$\ln\;(\frac{a^2b}c)\\$$
First we will use quotient rule then product rule and then at the end power rule of logarithms.
$$=\;\ln\;a^2b\;-\;\ln\;c\\\\=\;\ln\;a^2\;+\;\ln\;b\;-\;\ln\;c\\\\=\;2\;\ln\;a\;+\;\ln\;b\;-\;\ln\;c\\$$
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(iv) $$Log\;{(\frac{x\;y}z)}^\frac19$$
$$=\;\frac19Log\;(\frac{x\;y}z)\\=\;\frac19\;(Log\;xy\;-\;Log\;z)\\=\;\;\frac19\;(Log\;x\;+\;Log\;y\;-\;Log\;z)$$
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(v) $$Ln\;\sqrt[3]{16\;x^3}$$
$$=\frac13\;\ln\;(16\;x^3)\\\\=\frac13(\;\ln\;16\;+\;\ln\;x^3\;)\\\\=\frac13(\;\ln\;2^4\;+\;3\;\ln\;x\;)\\\\=\;\frac13\;(4\;\ln\;2\;+\;3\;\ln\;x\;)\\\\=\frac43\;\ln\;2\;+\;\ln\;x$$
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(vi) $$Log_2\;{(\frac{1-a}b)}^5\\\\=5\;\lbrack\;Log_2(1-a)\;-\;Log_2\;b\;\rbrack$$
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Q4 Find the value of x in the following equations:
(i) Log 2 + Log x = 1
Rule Log m n = Log m + Log n
Log 2x = 1
No base is given, so base will be 10.
When we shift common Log on the other side, right hand side will become the power of 10.
we can also say that Logarithmic form is converted into exponential form.
$$2x\;=\;10^1$$
2x = 10
x = 10/2 = 5
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(ii) $$Log_2x\;+\;Log_28\;=\;5$$
We will use product law of logarithms.
$$Log_2\;8x\;=\;5$$
Convert Logarithmic form into exponential form
$$2^5\;=\;8x$$
x = 32/8 = 4
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(iii) $${(81)}^x\;=\;{(243)}^{x+2}$$
$${(3^4)}^x\;=\;{(3^5)}^{x+2}\;$$
We will use rule of power of power.
$${(3)}^{4x}\;=\;{(3)}^{5x\;+\;10}\;$$
In equality if bases are same, powers will also be same.
4x = 5x + 10
4x – 5x = 10
-x = 10
x = -10
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(iv) $${(\frac1{27})}^{x-6}\;=27\\$$
Law of division of radicals will be used.
$$\frac{{(1)}^{x-6}}{{(27)}^{x-6}}\;=\;27\\$$
Any power of 1 will give us 1.
$$1\;=\;27\;{(27)}^{x-6}\\$$
Law of multiplication of indices will be used.
In multiplication if bases are same, powers will be added.
$$1\;=\;{(27)}^{1+x-6}\\$$
Any number raised to power zero will give us 1.
$$\;\;{(27)}^0=\;{(27)}^{x-5}\\$$
In equality if bases are same, powers will also be same.
0= x -5
x = 5
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(v) Log ( 5x-10 ) = 2
No base is given so base will be 10.
Convert Logarithmic form into exponential form.
$$\;\;{(10)}^2=\;5x\;-\;10\\\\$$
100 = 5x -10
100+10 = 5x
110 = 5x
110/5 = x
x = 22
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(vi) $$Log_2\;(x+1)\;-\;Log_2\;(x-4)\;=\;2\\$$
Quotient law of logarithms will be used.
$$Log_2\frac{x+1}{x-4}=2$$
Convert logarithmic form into exponential form.
$$2^2\;=\;\frac{x+1}{x-4}$$
4(x-4) = x+1
4x-16 = x+1
4x-x = 1 +16
3x = 17
x = 17/3
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Q5 Find the values of the following with the help of Logarithm table:
(i) $$\frac{3.68\;\times4.21}{5.234}\\$$
$$Let\;\;x\;=\;\frac{3.68\;\times4.21}{5.234}\\$$
Taking common log on both the sides
$$\log\;x\;=\;\log\;\frac{3.68\;\times4.21}{5.234}\\$$
Apply the laws of logarithms
Product and quotient law of logarithm will be used.
Log x = Log 3.68 + log 4.21 – log 5.234
= 0.5658 + 0.6243 – 0.7188
=0.4713
x = Antilog 0.4713
x= 2.960
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(ii) ( 4.67 ) ( 2.11 ) ( 2.397 )
Let x = ( 4.67 ) ( 2.11 ) ( 2.397 )
Taking log on both the sides
log x = log {( 4.67 ) ( 2.11 ) ( 2.397 ) }
Applying Log m n = log m + log n
Log x = Log 4.67 + Log 2.11 + Log 2.397
= 0.6693 + 0.3243 + 0.3797 = 1.3733
Taking Antilog on both the sides
x = Antilog 1.3733 = 23.62
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(iii) $$\frac{{(20.46)}^2\;\times\;(\;2.4122\;)}{754.3}$$
Let $$x\;=\;\frac{{(20.46)}^2\;\times\;(\;2.4122\;)}{754.3}$$
Taking log on both the sides
$$\log\;x\;=\log\;\frac{{(20.46)}^2\;\times\;(\;2.4122\;)}{754.3}$$
Applying the product and quotient law of logarithms.
$$Logx=Log{(20.46)}^2+\log2.4122-\log754.3$$
Applying the power law of logarithms.
Log x = 2 log(20.46) + 0.3824 – 2.8776
Logx = 2 (1.3109 ) + 0.3824 -2.8776
= 2.6218 + 0.3824 -2.8776
= 0.1266
Taking antilog on both the sides
x = Antilog 0.1266
= 1.339
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(iv) $$\frac{\sqrt[3]{9.364}\;\times\;21.64}{3.21}$$
Let $$x\;=\;\frac{\sqrt[3]{9.364}\;\times\;21.64}{3.21}$$
Taking log on both the sides
$$\log\;x\;=\;\log\;\frac{\sqrt[3]{9.364}\;\times\;21.64}{3.21}$$
Applying laws of product and division of logarithms.
$$\log x=\log\;\sqrt[3]{9.364}+\log21.64-\;\log3.21$$
Applying power law of logarithms.
$$\log x=\frac13\log\;9.364+\log21.64-\;\log3.21$$
$$\log x=\frac13(0.9715)+1.3353-0.5065$$
Log x = 0.3239 + 1.3353 – 0.5065
=1.1527
Taking antilog on both the sides
x = Antilog 1.1527
= 14.21
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Q6 $$The\;formula\;to\;measure\;the\\magnitude\;of\;earthquakes\;is\\given\;by\;M\;=\;Log_{10}(\frac A{A_0})\\If\;amplitude\;A\;is\;10000\;and\\refrence\;amplitude\;A_0\;is\;10.\\What\;is\;the\;magnitude\;of\\earthquake\;?$$
$$M\;=\;Log_{10}(\frac A{A_0})\\\\=\;Log_{10}\;(\frac{10000}{10})$$
After division, we get
$$=\;Log_{10}(1000)\;\\\\=Log_{10}{(10)}^3\\$$
After, applying the power rule of Logarithms, we get
$$=3\;Log_{10}10\\$$
we know that log of any number with the same base results 1.
Hence,
= 3 (1 ) = 3
So, the magnitude of earthquake will be 3.
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Q 7 Abdullah invested Rs 100,000 in a saving scheme and gains interest at the rate of 5 % per annum so that the total value of this investment after t years is Rs y. This is modelled by an equation y = 100,000 multiplied by 1.05 raised to power t, where t is greater than equal to 0. Find after how many years the investment will be double.
$$y\;=\;100,000\;{(1.05)}^t$$
Taking log on both the sides
$$Log\;y\;=\;Log\;\{\;100,000\;{(1.05)}^t\;\}$$
Applying Log m n = Log m + Log n
$$Log\;y\;=\;Log\;100,000\;+\;Log\;{(1.05)}^t$$
Applying power law of logarithms.
$$Log\;y\;=\;Log\;{(10)}^5\;+t\;Log\;{1.05}$$
Again applying power law of logarithms and putting the value of y.
$$Log\;200,000\;=5\;Log\;10\;+t\;Log\;{1.05}$$
No base is given hence log will be common log and we know that common log of 10 is 1.
$$Log(100000\times2)=5(1)+t\;(\;0.0212\;)$$
Applying Log m n = Log m + Log n
$$Log(100000)\;+\;Log\;2=5+t\;(\;0.0212\;)$$
After finding the logarithms, we get
$$5\;+\;0.3010\;=\;5\;+\;t\;(0.0212)$$
0.3010 = t ( 0.0212 )
0.3010 / 0.0212 = t
t = 14.2
Hence after 14 years the investment will be doubled.
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Q8 $$Huria\;is\;hyking\;up\;a\\mountain\;where\\temperature\;(T)\\decreases\;by\;3\%\\(\;or\;a\;factor\;of\;0.97\;)\\for\;every\;100\;meters\\gained\;in\;altitude.\\The\;initial\;temperature\\(T_i)\;at\;sea\;level\;is\;20^\circ C.\\U\sin g\;the\;formula\\T\;=\;T_i\;\times\;0.97^\frac h{100}\;,\;\\calculate\;the\\temperature\;at\;an\;\\altitude\;(h)\;of\;500\;meters.$$
$$T\;=\;T_i\;\times\;0.97^\frac h{100}$$
Taking log on both the sides
$$Log\;T\;=Log\;(\;T_i\;\times\;0.97^\frac h{100})$$
Applying Log m n = Log m + Log n
$$Log\;T\;=\;Log\;\;T_i\;+\;Log\;0.97^\frac h{100}$$
Applying power rule of logarithms.
$$Log\;T\;=\;Log\;\;T_i\;+\frac h{100}(Log\;0.97)$$
After putting the values, we get
$$Log\;T\;=\;Log\;20\;+\frac{500}{100}(Log\;0.97)$$
$$Log\;T\;=\;1.3010\;+5\;(\overline1.9868)$$
Multiply 5 with 1 bar and also with 0.9868
$$\;=\;1.3010\;+\;\overline5\;+\;4.934$$
Lets simplify
Log T = 1.3010 – 5 + 4.934 = 1.235
Taking antilog on both the sides
T = Antilog ( 1.235 )
Hence
$$T\;=\;17.18^\circ C$$
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