SELINA Solution Class 9 Chapter 8 Logarithms Exercise 8B

Question 1

Express in terms of log 2 and log 3 : log 36

Sol:

log 36 = log( 2 x 2 x 3 x 3 )
           = log( 2² x 3² )
           = log( 2² ) + log( 3² ) ....[ log_amn = log_am + log_an ]
           = 2log2 + 2log3         .....[ log_amⁿ = nlog_am ]

Question 1.2

Express in terms of log 2 and log 3:

log 144

Sol:

log 144

= log( 2 × 2 × 2 × 2 × 3 × 3 )
= log( 2⁴ x 3² )
= log( 2⁴ ) + log( 3² )  ....[ log_amn = log_am + log_an]
= 4log2 + 2log3          ....[ log_amⁿ = nlog_am ]

Question 1.3

Express in terms of log 2 and log 3 : log 4.5

SoL:

log 4.5

= log${\displaystyle \frac{45}{10}}$

= log${\displaystyle \frac{5\times 3\times 3}{5\times 2}}$

 = log${\displaystyle \frac{3^2}{2}}$

= log 3² - log 2   .....[ log_a${\displaystyle \frac{m}{n}}$ = log_am - log_an ]

= 2log3 - log 2   ......[ log_amⁿ = nlog_am ]

Question 1.4

Express in terms of log 2 and log 3 :
${\displaystyle \text{log}\frac{26}{51}-\text{log}\frac{91}{119}}$

Sol:

${\displaystyle \text{log}\frac{26}{51}-\text{log}\frac{91}{119}}$

= ${\displaystyle \text{log}\left(\frac{\frac{26}{51}}{\frac{91}{119}}\right) ....[{\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)]}$

= ${\displaystyle \text{log}\frac{26}{51}\times \frac{119}{91}}$

= ${\displaystyle \text{log}\frac{2\times 13}{3\times 17}\times \frac{7\times 17}{7\times 13}}$

= ${\displaystyle \text{log}\frac{2}{3}}$
= log 2 - log 3      ......[ log_a${\displaystyle \frac{m}{n}}$ = log_am - log_an ]

Question 1.5

Express in terms of log 2 and log 3 :
${\displaystyle \text{log}\frac{75}{16}-2\text{log}\frac{5}{9}+\text{log}\frac{32}{243}}$

Sol:

${\displaystyle \text{log}\frac{75}{16}-2\text{log}\frac{5}{9}+\text{log}\frac{32}{243}}$

= ${\displaystyle \text{log}\frac{75}{16}-\text{log}{\left(\frac{5}{9}\right)}^2+\text{log}\frac{32}{243}}$

= ${\displaystyle \text{log}\frac{75}{16}-\text{log}(\frac{5}{9}\times \frac{5}{9})+\text{log}\frac{32}{243}}$

= ${\displaystyle \text{log}\frac{75}{16}-\text{log}\frac{25}{81}+\text{log}\frac{32}{243}}$

= ${\displaystyle \text{log}\left(\frac{\frac{75}{16}}{\frac{25}{81}}\right) .....[{\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)]}$

= ${\displaystyle \text{log}\left(\frac{75}{16}\right)\times \left(\frac{81}{25}\right)+\log \left(\frac{32}{243}\right)}$

= ${\displaystyle \text{log}\frac{3\times 25}{16}\times \frac{81}{25}+\text{log}\frac{32}{243}}$

= ${\displaystyle \text{log}\frac{243}{16}+\text{log}\frac{32}{243}}$

= ${\displaystyle \text{log}(\frac{243}{16}\times \frac{32}{243})}$     .....[log_am + log_an = log_amn]

= ${\displaystyle \text{log}\frac{32}{16}}$
= log 2

Question 2.1

Express the following in a form free from logarithm :2 log x - log y = 1

Sol:

Consider the given equation 
2log x - log y = 1
⇒ logx2 - log y = 1
⇒ ${\displaystyle \text{log}\left(\frac{x^2}{y}\right)}$ = log 10

⇒  ${\displaystyle \frac{x^2}{y}=10}$

⇒  x² = 10y

Question 2.2

Express the following in a form free from logarithm:
2 log x + 3 log y = log a

Sol:

Consider the given equation 
2 log x + 3 log y = log a
⇒ logx² + logy³ = log a
⇒ log x²y³ = log a
⇒ x²y³ = a

Question 2.3

Express the following in a form free from logarithm:
a log x - b log y = 2 log 3

Sol:

Consider the given equation
alogx - blogy = 2log3
⇒ log x^a - logy^b = log3²
⇒ ${\displaystyle \text{log}\left(\frac{x^a}{y^b}\right)=\log 9}$
⇒  ${\displaystyle \frac{x^a}{y^b}=9}$
⇒  x^a = 9y^b

Question 3.1

Evaluate  the following without using tables : 
log 5 + log 8 - 2 log 2

Sol:

Consider the given expression
log 5 + log 8 - 2 log 2        .....[ nlog_am = log_amⁿ ]
= log 5 + log 8 x 8 - log 2² ......[ nlog_am = log_amⁿ ]
= log 5 x 8 - log 2²
= log 40 - log 4

= log${\displaystyle \frac{40}{4}}$                          .....[ log_am - log_an = log_a${\displaystyle \left(\frac{m}{n}\right)}$]
= log 10 
= 1

Question 3.2

Evaluate the following without using tables :
log 4 + ${\displaystyle \frac{1}{3}}$ log 125 - ${\displaystyle \frac{1}{5}}$log 32 

Sol:

Consider the given expression
log 4 + ${\displaystyle \frac{1}{3}\text{log 125}-\frac{1}{5}}$ log 32

= ${\displaystyle \log 4+{\log \left(125\right)}^{\frac{1}{3}}-{\log \left(32\right)}^{\frac{1}{5}}}$     ...[ nlog_am = log_amⁿ ]

= ${\displaystyle \log 4+{\log \left(5^3\right)}^{\frac{1}{3}}-{\log \left(2^5\right)}^{\frac{1}{5}}}$

= log 4 + log 5 - log 2
= log 4 x 5 - log 2                ....[ log_am + log_an = log_amn] 
= ${\displaystyle \text{log}\left(\frac{20}{2}\right) .....[{\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)]}$
= log 10
= 1

Question 3.3

Evaluate the following without using tables : 
log₁₀8 + log₁₀25 + 2 log₁₀3 - log₁₀18

Sol:

Consider the given expression
log₁₀8 + log₁₀25 + 2 log₁₀3 - log₁₀18
= log₁₀8 + log₁₀25 + log₁₀3² - log₁₀18      ....[ nlog_am = log_amⁿ ]
= log₁₀8 + log₁₀25 + log₁₀9 - log₁₀18
= log₁₀8 x 25 x 9 - log₁₀18             ....[ log_al + log_am + log_an = log_almn ]

= log₁₀1800 - log₁₀18         

= ${\displaystyle {\log }_{10}\left(\frac{1800}{18}\right)}$

= ${\displaystyle {\log }_{10}100}$                       .....[ ∵ log₁₀100 = 2]
= 2

Question 4

Prove that : ${\displaystyle 2\text{log}\frac{15}{18}-\text{log}\frac{25}{162}+\text{log}\frac{4}{9}=\log 2}$

Sol:

We need to prove that
${\displaystyle 2\text{log}\frac{15}{18}-\text{log}\frac{25}{162}+\text{log}\frac{4}{9}=\log 2}$

LHS = ${\displaystyle 2\text{log}\frac{15}{18}-\text{log}\frac{25}{162}+\text{log}\frac{4}{9}}$

= ${\displaystyle \text{log}{\left(\frac{15}{18}\right)}^2-\text{log}\left(\frac{25}{162}\right)+\text{log}\left(\frac{4}{9}\right)}$   ....[ nlog_am = log_amⁿ ]

= ${\displaystyle \text{log}[\left(\frac{15}{18}\right)\times \left(\frac{15}{18}\right)]-\text{log}\frac{25}{162}+\text{log}\frac{4}{9}}$

= ${\displaystyle \text{log}\left(\frac{15}{18}\right)\times \left(\frac{15}{18}\right)\times \left(\frac{4}{9}\right)-\text{log}\left(\frac{25}{162}\right) .....[{\log }_{a}m+{\log }_{a}n={\log }_a\left(mn\right)]}$

= ${\displaystyle \text{log}\frac{\left(\frac{15}{18}\right)\times \left(\frac{15}{18}\right)\times \frac{4}{9}}{\frac{25}{162}} .....[{\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)]}$

= ${\displaystyle \text{log}\left(\frac{15}{18}\right)\times \left(\frac{15}{18}\right)\times \frac{4}{9}\times \frac{162}{25}}$

= ${\displaystyle \text{log}\frac{72}{36}}$
= log 2
= R.H.S.

Question 5

Find x, if : x - log 48 + 3 log 2 = ${\displaystyle \frac{1}{3}}$log 125 - log 3.

Sol:

Consider the given equation
x - log 48 + 3log2 = ${\displaystyle \frac{1}{3}}$log 125 - log 3

⇒ x = ${\displaystyle \frac{1}{3}}$log125 - log 3 + log 48 - 3 log 2

⇒ x = ${\displaystyle \text{log}{\left(125\right)}^{\frac{1}{3}}-\log 3+\log 48-{\log 2}^3 ....[n{\log }_{a}m={\log }_a{m}^n]}$

⇒ x = ${\displaystyle {\log (5\times 5\times 5)}^{\frac{1}{3}}-\log 3+\log 48-\log 8}$

⇒ x = ${\displaystyle {\log \left(5^3\right)}^{\frac{1}{3}}-\log 3+\log 48-\log 8}$

⇒ x = log 5 - log 3 + log 48 - log 8

⇒ x = log 5 + log 48 - log 3 - log 8

⇒ x = ( log 5 + log 48 ) - ( log 3 + log 8 )

⇒ x = ( log 5 x 48 ) - ( log 3 x 8 )        ....[ log_am + log_an = log_amn ]

⇒ x = log${\displaystyle \frac{5\times 48}{3\times 8} .....[{\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)]}$

⇒ x = ${\displaystyle \text{log}\frac{5\times 6\times 8}{3\times 8}}$

⇒ x = log 10

⇒ x = 1.

Question 6

Express log₁₀2 + 1 in the form of log₁₀x .

Sol:

log₁₀2 + 1

= log₁₀2 + log₁₀10         ....[ ∵ log₁₀10 = 1 ]

= log₁₀2 x 10                ....[ log_am + log_an = log_amn ]

= log₁₀20.

Question 7.1

Solve for x : log₁₀ (x - 10) = 1

Sol:

log₁₀  ( x - 10 ) = 1
⇒ log₁₀ ( x - 10 ) = log₁₀10
⇒ x - 10 = 10
⇒ x = 10 + 10
⇒ x = 20.

Question 7.2

Solve for x : log (x² - 21) = 2.

Sol:

log ( x² - 21 ) = 2
⇒ log ( x²  - 21) = log 100
⇒ x² - 21 - 100 = 0 
⇒ x² - 121 = 0 
⇒ x²  = 121
⇒ x = ${\displaystyle \pm \sqrt{121}}$
⇒ x =${\displaystyle \pm }$11

Question 7.3

Solve for x :  log (x - 2) + log (x + 2) = log 5

Sol:

log ( x - 2 ) + log ( x + 2 ) = log 5
⇒ log ( x - 2 ) ( x + 2) = log 5       ...[ log_a m + log_a n = log_a mn ]
⇒ log ( x² - 4 ) = log 5
⇒ x² -  4 = 5 
⇒ x ²  = 9 

⇒ x = ${\displaystyle \pm \sqrt{9}}$

⇒ x = ${\displaystyle \pm \sqrt{3^2}}$

⇒ x = ${\displaystyle \pm }$3 

Question 7.4

Solve for x : log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3

Sol:

log ( x + 5 ) + log ( x - 5 ) = 4log2 + 2log3

⇒ log ( x + 5 ) ( x - 5 ) = 4log 2 + 2log3  ...[ log_am + log_a n + log_a mn]

⇒ log ( x² - 25 ) = log2⁴  + log3²            ... [ n log_a m = log_a mⁿ ]

⇒ log ( x² - 25 ) = log 16 + log9

⇒ log ( x² - 25 )= log 16 x 9           ...[ log_a m + log_a n + log_a mn]  

⇒ log ( x²  - 25 ) = log 144

⇒ x² - 25 = 144

⇒ x²  = 144  + 25 

⇒ x²  = 169

⇒ x = ${\displaystyle \pm \sqrt{169}}$

⇒ x =${\displaystyle \pm \sqrt{{13}^2}}$

⇒ x =${\displaystyle \pm }$ 13

Question 8.1

log ( x + 5 ) + log ( x - 5 ) = 4log2 + 2log3

⇒ log ( x + 5 ) ( x - 5 ) = 4log 2 + 2log3  ...[ log_am + log_a n + log_a mn]

⇒ log ( x² - 25 ) = log2⁴  + log3²            ... [ n log_a m = log_a mⁿ ]

⇒ log ( x² - 25 ) = log 16 + log9

⇒ log ( x² - 25 )= log 16 x 9           ...[ log_a m + log_a n + log_a mn]  

⇒ log ( x²  - 25 ) = log 144

⇒ x² - 25 = 144

⇒ x²  = 144  + 25 

⇒ x²  = 169

⇒ x = ${\displaystyle \pm \sqrt{169}}$

⇒ x =${\displaystyle \pm \sqrt{{13}^2}}$

⇒ x =${\displaystyle \pm }$ 13

Sol:

${\displaystyle \frac{\log 81}{\log 27}}$ = x 

⇒ x = ${\displaystyle \frac{\log 81}{\log 27}}$

⇒ x = ${\displaystyle \frac{\log 3\times 3\times 3\times 3}{\log 3\times 3\times 3}}$

⇒ x = ${\displaystyle \frac{{\log 3}^4}{{\log 3}^3}}$

⇒ x = ${\displaystyle \frac{4\log 3}{3\log 3}}$   ... [ n log_am = log_amⁿ ]

⇒ x= ${\displaystyle \frac{4}{3}}$

⇒ x = 1 ${\displaystyle \frac{1}{3}}$

Question 8.2

Solve for x : ${\displaystyle \frac{\log 128}{\log 32}}$ = x

Sol:

 ${\displaystyle \frac{\log 128}{\log 32}}$ = x

${\displaystyle \Rightarrow x=\frac{\log 128}{\log 32}}$

${\displaystyle \Rightarrow x=\frac{\log 2\times 2\times 2\times 2\times 2\times 2\times 2}{\log 2\times 2\times 2\times 2\times 2}}$

${\displaystyle \Rightarrow x=\frac{{\log 2}^7}{{\log 2}^5}}$

${\displaystyle \Rightarrow x=\frac{7\log 2}{5\log 2}}$     ... [ n log_a m = log_a mⁿ  ]

${\displaystyle \Rightarrow x=\frac{7}{5}}$

⇒ x = 1.4

Question 8.3

Solve for x : ${\displaystyle \frac{\log 64}{\log 8}}$ = log x

Sol:

⇒ ${\displaystyle \frac{\log 64}{\log 8}}$ = log x 

⇒ log x = ${\displaystyle \frac{\log 64}{\log 8}}$ 

⇒ log x = ${\displaystyle \frac{\log 2\times 2\times 2\times 2\times 2\times 2}{\log 2\times 2\times 2}}$

⇒ log x =${\displaystyle \frac{{\log 2}^6}{{\log 2}^3}}$  

⇒ log x = ${\displaystyle \frac{6\log 2}{3\log 2}}$ ... [ n log_a m = log_a mⁿ ] 

⇒ log x = ${\displaystyle \frac{6}{3}}$

⇒ log x = 2 

⇒ log₁₀ x = 2 

⇒ 10² = x 

⇒ x = 10 x 10 

⇒ x = 100

Question 9

Given that log x = m + n and log y = m - n, express the value of log ${\displaystyle \frac{10x }{y^2 }}$  in terms of m and n.

Sol:

 Given that
log x = m + n ; 
log y = m - n ;
Consider the expression log ${\displaystyle \frac{10x}{y^2}}$ : 

log ${\displaystyle \frac{10x }{y^2 }}$

= log 10 x - log y² 
⇒ log 10 x - 2 log y  ... [ n log_a m = log_a mⁿ]
⇒ log 10 + log x - 2 log y ...[ log_a m + log_a n = log_a mn ]
⇒ 1 +  log x - 2 log y   
⇒  1 + m + n - 2 ( m - n )
⇒  1 + m + n - 2m + 2n
⇒ log ${\displaystyle \frac{10x }{y^2}}$ = 1 - m + 3n.

Question 10.1

State, true or false : log 1 x log 1000 = 0

Sol:

We have,
log 1 = 0 and log 1000 = 3
∴ log 1 x log 1000 = 0 x 3 = 0
Thus the statement, log 1 x log 1000 = 0 is true.

Question 10.2

State, true or false : 
${\displaystyle \frac{\log x}{\log y}}$ = log x - log y

Sol:

We know that
${\displaystyle \log \left(\frac{m}{n}\right)=\log m-\log n}$
∴ ${\displaystyle \frac{\log x}{\log y}\ne \log x-\log y}$
Thus the statement, ${\displaystyle \frac{\log x}{\log y}}$= log x - log y is false.

Question 10.3

State, true or false : 
If ${\displaystyle \frac{\log 25}{\log 5}=\log x}$, then x = 2.

Sol:

Given that
${\displaystyle \frac{\log 25}{\log 5}}$ = log x

⇒ ${\displaystyle \frac{\log 5\times 5}{\log 5}}$ = log x

⇒ ${\displaystyle \frac{{\log 5}^2}{\log 5}=\log x}$

⇒ ${\displaystyle \frac{2\log 5}{\log 5}=\log x ...[{\log }_a{m}^n=n{\log }_{a}m]}$

⇒ 2 = log₁₀x
⇒ 10² = x
⇒ x = 100
Thus, the statement, x = 2 is false.

Question 10.4

State, true or false : 
log x x log y = log x + log y

Sol:

We know that
logx + log y = logxy
∴ logx + log y ≠ logx x log y
Thus the statement logx + log y = logx x log y is false.

Question 11.1

If log₁₀2 = a and log₁₀3 = b ; express each of the following in terms of 'a' and 'b': log 12

Sol:

Given that log₁₀2 = a and log₁₀3 = b
log 12
= log 2 x 2 x 3
= log 2 x 2 + log 3       ...[ log_amn = log_am + log_an ]
= log 2² + log 3
= 2log 2 + log 3          ...[ nlog_am = log_amⁿ ]
= 2a + b                      ...[ ∵ log₁₀2 = a and log₁₀3 = b ]

Question 11.2

If log₁₀2 = a and log₁₀3 = b ; express each of the following in terms of 'a' and 'b': log 2.25

Sol:

log₁₀2 = a and log₁₀3 = b
log 2.25

= log${\displaystyle \frac{225}{100}}$

= log${\displaystyle \frac{25\times 9}{25\times 4}}$

= log${\displaystyle \frac{25\times 9}{25\times 4}}$

= log${\displaystyle \left(\frac{9}{4}\right)}$

= ${\displaystyle {\log \left(\frac{3}{2}\right)}^2}$

= ${\displaystyle \log \left(\frac{3}{2}\right)}$                 ...[ nlog_am = log_amⁿ ]

= 2( log3 - log2 )        ...[ log_am - log_an = log_a${\displaystyle \left(\frac{m}{n}\right)}$]
= 2( b - a )                  ...[ ∵ log₁₀2 = a and log₁₀3 = b ]
= 2b - 2a

Question 11.3

log₁₀2 = a and log₁₀3 = b
log 2.25

= log${\displaystyle \frac{225}{100}}$

= log${\displaystyle \frac{25\times 9}{25\times 4}}$

= log${\displaystyle \frac{25\times 9}{25\times 4}}$

= log${\displaystyle \left(\frac{9}{4}\right)}$

= ${\displaystyle {\log \left(\frac{3}{2}\right)}^2}$

= ${\displaystyle \log \left(\frac{3}{2}\right)}$                 ...[ nlog_am = log_amⁿ ]

= 2( log3 - log2 )        ...[ log_am - log_an = log_a${\displaystyle \left(\frac{m}{n}\right)}$]
= 2( b - a )                  ...[ ∵ log₁₀2 = a and log₁₀3 = b ]
= 2b - 2a

Sol:

Given that log₁₀2 = a and log₁₀3 = b
log ${\displaystyle 2\frac{1}{4}}$
= log ${\displaystyle \left(\frac{9}{4}\right)}$
= log ${\displaystyle {\left(\frac{3}{2}\right)}^2}$         
= 2log${\displaystyle \left(\frac{3}{2}\right)}$                ...[ nlog_am = log_amⁿ ]
= 2( log3 - log2 )         ...[ log_am - log_an = log_a ${\displaystyle \frac{m}{n}}$ ]
= 2( b - a )                   ...[ ∵ log₁₀2 = a and log₁₀3 = b ]
= 2b - 2a

Question 11.4

If log₁₀2 = a and log₁₀3 = b ; express each of the following in terms of 'a' and 'b' : log 5.4 

Sol:

Given that log₁₀2 = a and log₁₀3 = b
log 5.4 
= log${\displaystyle \frac{54}{10}}$
= log${\displaystyle \frac{2\times 3\times 3\times 3}{10}}$
= log( 2 x 3 x 3 x 3 ) - log₁₀10   ...[ log_am = log_an = log_a${\displaystyle \left(\frac{m}{n}\right)}$ ]
= log₁₀2 + log₁₀3³ - log₁₀10   ...[log_amn = log_am + log_an]
= log₁₀2 + 3log₁₀3 - log₁₀10  ...[ nlog_am = log_amⁿ ]
= log₁₀2 + 3log₁₀3 - 1           ...[ ∵ log₁₀10 = 1]
= a + 3b - 1                            ...[ ∵ log₁₀2 = a and log₁₀3 = b ]

Question 11.5

If log₁₀2 = a and log₁₀3 = b; express each of the following in terms of 'a' and 'b' : log 60

Sol:

log 60 
= log₁₀10 x 2 x 3           ...[ log_amn = log_am + log_an ]
= 1 + log₁₀2 + log₁₀3  ...[ ∵ log1010 = 1 ] 
= 1 + a + b                   ...[ ∵ log₁₀2 = a and log₁₀3 = b ]

Question 11.6

If log₁₀2 = a and log₁₀3 = b; express each of the following in terms of 'a' and 'b' :  log ${\displaystyle 3\frac{1}{8}}$

Sol:

log ${\displaystyle 3\frac{1}{8}}$

= ${\displaystyle {\log }_{10}(\frac{25}{8}\times \frac{4}{4})}$

= ${\displaystyle {\log }_{10}\left(\frac{100}{32}\right)}$

= ${\displaystyle {\log }_{10}100-{\log }_{10}32 ...[{\log }_a\left(\frac{m}{n}\right)={\log }_{a}m-{\log }_{a}n]}$
= ${\displaystyle {\log }_{10}100-{\log }_{10}{2}^5}$
= 2 - log₁₀2⁵              ...[ ∵ log₁₀100 = 2 ]
= 2 - 5log₁₀2             ...[ log_amⁿ = nlog_am ]
= 2 - 5a                     ...[ ∵ log₁₀2 = a ]

Question 12.1

If log 2 = 0.3010 and log 3 = 0.4771 ;  find the value of : log 12

Sol:

We know that log 2 = 0.3010 and log 3 = 0.4771
log 12
= log 2 x 2 x 3
= log 2 x 2 + log 3           ...[ log_amn = log_am + log_an ]
= log2² + log3
= 2log2 + log 3                ...[ nlog_am = log_amⁿ ]
= 2( 0.3010 ) + 0.4771      ...[ ∵ log 2 = 0.3010 and log3 = 0.4771 ]

= 1.0791

Question 12.2

If log 2 = 0.3010 and log 3 = 0.4771 ; find the value of : log 1.2

Sol:

log 2 = 0.3010 and log 3 = 0.4771
log 1.2
= log${\displaystyle \frac{12}{10}}$
= log 12 - log 10            ...[ loga${\displaystyle \frac{m}{n}}$ = log_am - log_an ]
= log 2 x 2 x 3 - 1           ...[ ∵ log 10 = 1 ]
= log 2 x 2 + log 3 - 1    ...[ log_amn = log_am + log_an ]
= log 22 + log 3 - 1
= 2log2 + log3 - 1          ...[ nlogam = logamn ]
= 2( 0.3010 ) + 0.4771 - 1  ...[ ∵ log 2 = 0.3010 and log3 = 0.4771 ] 
= 1.0791 - 1
= 0.0791

Question 12.3

If log 2 = 0.3010 and log 3 = 0.4771; find the value of : log 3.6

Sol:

We know that log 2 = 0.3010 and log 3 = 0.4771.
log 3.6
= log${\displaystyle \frac{36}{10}}$
= log 36 - log 10       ...${\displaystyle [{\log }_a\left(\frac{m}{n}\right)={\log }_{a}m-{\log }_{a}n]}$
= log 2 x 2 x 3 x 3 - 1    ...[ ∵ log 10 = 1 ]
= log 2 x 2 + log 3 x 3 - 1  ...${\displaystyle [{\log }_{a}mn={\log }_{a}m+{\log }_{a}n]}$
= log 2² + log 3² - 1    ]
= 2log2 + 2log3 - 1             ...${\displaystyle [n{\log }_{a}m={\log }_a{m}^n]}$ 
= 2(0.3010) + 2(0.4771) - 1
= 1.5562 - 1                       ...${\displaystyle [\because \log 2=0.3010\text{and}\log 3=0.4771]}$
= 0.5562    

Question 12.4

If log 2 = 0.3010 and log 3 = 0.4771; find the value of : log 15

Sol:

We know that log 2 = 0.3010 and log 3 = 0.4771.
log 15
= log${\displaystyle (\frac{15}{10}\times 10)}$
= log${\displaystyle \left(\frac{15}{10}\right)}$ + log 10
= log${\displaystyle \left(\frac{3}{2}\right)}$ + 1                   ...[ ∵ log 10 = 1 ]
= log 3 - log 2 + 1                ...${\displaystyle [\because \log m-\log n=\log \left(\frac{m}{n}\right)]}$
= 0.4771 - 0.3010 + 1
= 1.1761

Question 12.5

If log 2 = 0.3010 and log 3 = 0.4771; find the value of : log 25

Sol:

We know that log 2 = 0.3010 and log 3 = 0.4771
log 25
= log${\displaystyle (\frac{25}{4}\times 4)}$
= log${\displaystyle \left(\frac{100}{4}\right)}$           ...${\displaystyle [{\log }_{a}mn={\log }_{a}m+{\log }_{a}n]}$
= log 100 - log( 2 x 2 )   ...${\displaystyle [{\log }_a\left(\frac{m}{n}\right)={\log }_{a}m-{\log }_{a}n]}$
= 2 - log(2²)                   ...[ log 100 = 2 ]
= 2 - 2log2                     ...${\displaystyle [{\log }_a{m}^n=n{\log }_{a}m]}$
= 2 - 2( 0.3010 )             ...[ ∵ log 2 = 0.3010 ]
= 1.398

Question 12.6

If log 2 = 0.3010 and log 3 = 0.4771; find the value of : ${\displaystyle \frac{2}{3}}$ log 8

Sol:

We know that log 2 = 0.3010 and log 3 = 0.4771
${\displaystyle \frac{2}{3}}$log 8
= ${\displaystyle \frac{2}{3}}$ log 2 x 2 x 2
= ${\displaystyle \frac{2}{3}}$ log2³ 
= 3 x ${\displaystyle \frac{2}{3}}$ log 2               ...[ log_amⁿ = nlog_am ]
= 2 log 2
= 2 x 0.3010                   ...[ ∵ log 2 = 0.3010 ]
= 0.602

Question 13

Given 2 log₁₀ x + 1 = log₁₀ 250, find :
(i) x
(ii) log₁₀ 2x

Sol:

(i) Consider the given equation : 
2log₁₀x + 1 = log₁₀250
⇒ log₁₀x² + 1 = log₁₀250     [ log_amⁿ = nlog_am]

⇒ log₁₀x2 + log₁₀10 = log₁₀250  [ ∵ log₁₀10 = 1]

⇒ log₁₀( x² x 10 ) = log₁₀250       [ log_am + log_an = log_amn ]

⇒ x² x 10 = 250
⇒ x2 = 25
⇒ x = ${\displaystyle \sqrt{25}}$
⇒ x = 5

(ii) x = 5 ( proved above in (i))
log₁₀2x = log₁₀2(5)
= log₁₀10
= 1                               [ ∵ log₁₀10 = 1]

Question 14

Given 3log x + ${\displaystyle \frac{1}{2}}$log y = 2, express y in term of x.

Sol:

3log x + ${\displaystyle \frac{1}{2}}$log y = 2
⇒ log x³ + log√y = 2
⇒ log x³√y  = 2
⇒  x³√y = 10²
⇒ √y = ${\displaystyle \frac{{10}^2}{x^3}}$
Squaring both sides, we get
y = ${\displaystyle \frac{10000}{x^6}}$

⇒ y = ${\displaystyle 10000x^{-6}}$

Question 15

If x = (100)^a , y = (10000)^b and z = (10)^c , find log${\displaystyle \frac{10\sqrt{y}}{x^2{z}^3}}$ in terms of a, b and c.

Sol:

x = (100)^a , y = (10000)^b and z = (10)^c

⇒ log x = alog 100, log y = b log 10000 and log z = clog 10

${\displaystyle \text{log}\frac{10\sqrt{y}}{x^2{z}^3}}$ = log10√y - log( x²z³ )

= log( 10y^1/2 ) - logx² - logz³

= log 10 + logy^1/2 - logx² - logz³
= log 10  + ${\displaystyle \frac{1}{2}}$log y - 2log x - 3log z

= 1 + ${\displaystyle \frac{1}{2}{\log \left(10000\right)}^b-2{\log \left(100\right)}^a-3{\log \left(10\right)}^c}$       ....(Since log 10 = 1 )

= 1 + ${\displaystyle \frac{b}{2}{\log \left(10\right)}^4-a{\log \left(10\right)}^2-3c\log 10}$

= 1 + ${\displaystyle \frac{b}{2}\times 4\log 10-2\times 2a\log 10-3c\log 10}$

= 1 + 2b - 4a - 3c

Question 16

If 3( log 5 - log 3 ) - ( log 5 - 2 log 6 ) = 2 - log x, find x.

Sol:

3( log 5 - log 3 ) - ( log 5 - 2 log 6 ) = 2 - log x

⇒ 3log 5 - 3log 3 - log 5 + 2 log( 2 x 3 ) = 2 - log x
⇒  3log 5 - 3log 3 - log 5 + 2log2 + 2log3 = 2 - log x
⇒ 2log5 - log3 + 2log2 = 2 - log x
⇒ 2log5 - log3 + 2log2 + log x = 2
⇒ log5² - log 3 + log2² + log x = 2

⇒ log${\displaystyle \left(\frac{25\times 4\times x}{3}\right)=2}$ 

⇒ log${\displaystyle \left(\frac{100x}{3}\right)=2}$

⇒ ${\displaystyle \frac{100x}{3}={10}^2}$

⇒ ${\displaystyle \frac{x}{3}=1}$
⇒ x = 3.