SELINA Solution Class 9 Chapter 8 Logarithms Exercise 8C

Question 1.1

If log₁₀ 8 = 0.90; find the value of : log₁₀ 4

Sol:

Given that log₁₀8 = 0.90

⇒ log₁₀2 x 2 x 2 = 0.90

⇒ log₁₀2³ = 0.90

⇒ 3log₁₀2 = 0.90

⇒  log₁₀2 = ${\displaystyle \frac{0.90}{3}}$

⇒  log₁₀2  = 0.30            ...(1)

log 4
= log₁₀( 2 x 2 )
= log₁₀( 2² )
= 2log₁₀2 
= 2( 0.30 )                ...[ from(1) ]
= 0.60

Question 1.2

If log₁₀ 8 = 0.90; find the value of : log√32

Sol:

Given that log₁₀8 = 0.90

⇒ log₁₀2 x 2 x 2 = 0.90

⇒ log₁₀2³ = 0.90

⇒ 3log₁₀2 = 0.90

⇒  log₁₀2 = ${\displaystyle \frac{0.90}{3}}$

⇒  log₁₀2  = 0.30            ...(1)

log √32
=${\displaystyle {\log }_{10}{\left(32\right)}^{\frac{1}{2}}}$

= ${\displaystyle \frac{1}{2}{\log }_{10}\left(32\right)}$

= ${\displaystyle \frac{1}{2}{\log }_{10}(2\times 2\times 2\times 2\times 2)}$

= ${\displaystyle \frac{1}{2}{\log }_{10}\left(2^5\right)}$

= ${\displaystyle \frac{1}{2}\times 5{\log }_{10}2}$

= ${\displaystyle \frac{1}{2}\times 5\left(0.30\right)}$           [ from(1) ] 

= 5 x 0.15
= 0.75

Question 1.3

Given that log₁₀8 = 0.90

⇒ log₁₀2 x 2 x 2 = 0.90

⇒ log₁₀2³ = 0.90

⇒ 3log₁₀2 = 0.90

⇒  log₁₀2 = ${\displaystyle \frac{0.90}{3}}$

⇒  log₁₀2  = 0.30            ...(1)

log √32
=${\displaystyle {\log }_{10}{\left(32\right)}^{\frac{1}{2}}}$

= ${\displaystyle \frac{1}{2}{\log }_{10}\left(32\right)}$

= ${\displaystyle \frac{1}{2}{\log }_{10}(2\times 2\times 2\times 2\times 2)}$

= ${\displaystyle \frac{1}{2}{\log }_{10}\left(2^5\right)}$

= ${\displaystyle \frac{1}{2}\times 5{\log }_{10}2}$

= ${\displaystyle \frac{1}{2}\times 5\left(0.30\right)}$           [ from(1) ] 

= 5 x 0.15
= 0.75

Sol:

Given that log₁₀8 = 0.90

⇒ log₁₀2 x 2 x 2 = 0.90

⇒ log₁₀2³ = 0.90

⇒ 3log₁₀2 = 0.90

⇒  log₁₀2 = ${\displaystyle \frac{0.90}{3}}$

⇒  log₁₀2  = 0.30            ...(1)

log 0.125

= log₁₀${\displaystyle \frac{125}{1000}}$

= log₁₀${\displaystyle \frac{1}{8}}$

= ${\displaystyle {\log }_{10}\left(\frac{1}{2\times 2\times 2}\right)}$

= ${\displaystyle {\log }_{10}\left(\frac{1}{2^3}\right)}$

= log₁₀2^-3 
= - 3 x ( 0.30 )          [ from(1) ]
= - 0.9

Question 2.1

If log 27 = 1.431, find the value of : log 9

Sol:

log 27 = 1.431
⇒ log 3 x 3 x 3 = 1.431
⇒ log 3³ = 1.431
⇒ 3log3 = 1.431
⇒ log 3 = ${\displaystyle \frac{1.431}{3}}$
⇒ log 3 = 0.477              ...(1)

log 9
= log( 3 x 3 )
= log 32
= 2 log 3
= 2 x 0.477                   ...[ from(1) ]
= 0.954

Question 2.2

If log 27 = 1.431, find the value of : log 300

Sol:

log 27 = 1.431
⇒ log 3 x 3 x 3 = 1.431
⇒ log 3³ = 1.431
⇒ 3log3 = 1.431
⇒ log 3 = ${\displaystyle \frac{1.431}{3}}$
⇒ log 3 = 0.477              ...(1)

log 300
= log( 3 x 100 )
= log 3 + log 100
= log 3 + 2              ...[ ∵ log₁₀100 = 2 ]
= 0.477 + 2
= 2.477

Question 3

If log₁₀ a = b, find 10^{3b - 2} in terms of a.

Sol:

log₁₀ a = b
⇒ 10^b = a

⇒ ( 10^b )³ = a³    ...[ Cubing both sides ]

⇒ ${\displaystyle \frac{{10}^{3b}}{{10}^2}=\frac{a^3}{{10}^2}}$  ...[dividing both sides by 10²]

⇒ 10^{3b - 2} = ${\displaystyle \frac{a^3}{100}}$

Question 4

If log₅ x = y, find 5^{2y+ 3} in terms of x.

Sol:

log₅ x = y    ...[ given ]

⇒ 5^y = x

⇒ (5^y)² = x²

⇒  5^2y = x²

⇒ 5^2y x 5³ = x² x 5³

 ⇒ 5^{2y + 3} = 125x²

Question 5.1

Given: log₃ m = x and log₃ n = y.
Express 3^{2x - 3} in terms of m.

Sol:

Given that log₃m = x and log₃n = y
⇒ 3^x = m and 3^y = n 

Consider the given expression :
3^{2x - 3}
= 3^2x . 3^-3
=${\displaystyle 3^{2x}.\frac{1}{3^3}}$

= ${\displaystyle \frac{3^{2x}}{3^3}}$

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

= ${\displaystyle \frac{m^2}{27}}$
Therefore, 3^{2x - 3} = ${\displaystyle \frac{m^2}{27}}$

Question 5.2

Given: log₃ m = x and log₃ n = y.
Write down 3^{1 - 2y + 3x} in terms of m and n.

Sol:

Given that log₃m = x and log₃n = y
⇒ 3^x = m and 3^y = n 
Consider the given expression :
${\displaystyle 3^{1-2y+3x}=3^1.3^{-2y}.3^{3x}}$

= 3 . ${\displaystyle \frac{1}{3^{2y}}.3^{3x}}$ 

= ${\displaystyle \frac{3}{{\left(3^y\right)}^2}.{\left(3^x\right)}^3}$

= ${\displaystyle \frac{3}{{\left(n\right)}^2}.{\left(m\right)}^3}$

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

Therefore, ${\displaystyle 3^{1-2y+3x}=\frac{{\left(3m\right)}^3}{n^2}}$

Question 5.3

Given: log₃ m = x and log₃ n = y.
If 2 log₃ A = 5x - 3y; find A in terms of m and n.

Sol:

Given that log₃m = x and log₃n = y
⇒ 3^x = m and 3^y = n 

Consider the given expression :
2log₃A = 5x - 3y
⇒ 2log₃A = 5 log₃m - 3log₃n

⇒ log₃A² = log₃m⁵ - log₃n³

⇒ log₃A² = ${\displaystyle {\log }_3\left(\frac{m^5}{n^3}\right)}$

⇒ A² = ${\displaystyle \left(\frac{m^5}{n^3}\right)}$

⇒ A = ${\displaystyle \sqrt{\left(\frac{m^5}{n^3}\right)}}$

Question 6.1

Simplify : log (a)³ - log a

Sol:

log (a)³ - log a 
= 3 log a - log a
= 2 log a

Question 6.2

Simplify : log (a)³ ÷  log a

Sol:

log (a)³ ÷  log a
= 3 log a ÷  log a
= ${\displaystyle \frac{3\log a}{\log a}}$
= 3

Question 7

If log (a + b) = log a + log b, find a in terms of b.

Sol:

log  ( a + b ) = log a + log b

⇒ log ( a + b ) = log ab

⇒ a + b= ab

⇒ a - ab = - b

⇒ - ab + a = - b

⇒ - a ( b - 1 ) = - b 

⇒ a ( b -1 ) = b

⇒ a = ${\displaystyle \frac{b}{b-1}}$

Question 8.1

Prove that :  (log a)² - (log b)² = log ${\displaystyle \left(\frac{a}{b}\right)}$ . Log (ab)

Sol:

L.H.S = ( log a )²  - ( log b  )² 

⇒ L.H.S = ( log a + log b ) ( log a - log b )

⇒ L.H.S = log ( ab ) log ${\displaystyle \left(\frac{a}{b}\right)}$

⇒ L.H.S = log ${\displaystyle \left(\frac{a}{b}\right)}$ x log ( ab) 

⇒ L.H.S = R.H.S

Hence proved.

Question 8.2

Prove that :  If a log b + b log a - 1 = 0, then b^a. a^b = 10

Sol:

Given that

a log b + b log a - 1 = 0

⇒ a log b + b log a  = 1 

⇒ log b^a + loga^b =1

⇒ log b^a + log a^b = log 10

⇒ log ( b^a . a^b ) = log 10

⇒ b^a . a^b = 10

Question 9.1

 If log (a + 1) = log (4a - 3) - log 3; find a.

Sol:

Given that

 log (a + 1) = log (4a - 3) - log 3 

⇒ log (a + 1) = log ${\displaystyle \left(\frac{4a-3}{3}\right)}$

⇒ a + 1 = ${\displaystyle \frac{4a-3}{3}}$

⇒ 3a + 3 = 4a - 3

⇒ 4a - 3a = 3 + 3

⇒ a = 6

Question 9.2

If 2 log y - log x - 3 = 0, express x in terms of y.

Sol:

2 log y - log x - 3 = 0

⇒ 2 log y - log x = 3

⇒ log y² - log x = 3

⇒ log y² - log x = log 1000

⇒ log ${\displaystyle \frac{y^2}{x }}$  = log 1000

⇒ ${\displaystyle \frac{y^2}{x }}$ = 1000

⇒ x = ${\displaystyle \frac{y^2}{1000}}$

Question 9.3

2 log y - log x - 3 = 0

⇒ 2 log y - log x = 3

⇒ log y² - log x = 3

⇒ log y² - log x = log 1000

⇒ log ${\displaystyle \frac{y^2}{x }}$  = log 1000

⇒ ${\displaystyle \frac{y^2}{x }}$ = 1000

⇒ x = ${\displaystyle \frac{y^2}{1000}}$

Sol:

log₁₀ 125 = 3 ( 1 - log₁₀2)

L.H.S.  = log₁₀ 125 

⇒ log₁₀ 5 x 5 x 5

⇒ log₁₀ 5³

⇒ 3 log₁₀ 5        ....( 1 )

 R.H.S = 3 ( 1 - log₁₀ 2 )

⇒ 3 log₁₀ 10 - log₁₀ 2 )

⇒ 3 log₁₀  ${\displaystyle \left(\frac{10}{2}\right)}$ 

⇒ 3 log₁₀ 5        ......( 2 )

From ( 1 ) and ( 2 ) , we have

L.H.S. = R. H. S.

Hence proved.

Question 10

Given log x = 2m - n , log y = n - 2m and log z = 3m - 2n , find in terms of m and n, the value of log ${\displaystyle \frac{x^2{y}^3}{z^4}}$.

Sol:

Given log x = 2m - n, log y = n - 2m, log z = 3m - 2n.

Given : log ${\displaystyle \frac{x^2{y}^3}{z^4}}$

We know that log(a/b) = log a - log b.

⇒ log ${\displaystyle \left(x^2{y}^3\right)}$ - log ${\displaystyle \left(z^4\right)}$

We know that log(ab) = log a + log b

⇒ log ${\displaystyle \left(x^2\right)}$ + log ${\displaystyle \left(y^3\right)}$ - log ${\displaystyle \left(z^4\right)}$

⇒ 2 log x + 3 log y - 4 log z

⇒ 2(2m - n) + 3(n - 2m) - 4(3m - 2n)

⇒ 4m - 2n + 3n - 6m - 12m + 8n

⇒ -14m + 9n

Question 11

Given log x = 2m - n, log y = n - 2m, log z = 3m - 2n.

Given : log ${\displaystyle \frac{x^2{y}^3}{z^4}}$

We know that log(a/b) = log a - log b.

⇒ log ${\displaystyle \left(x^2{y}^3\right)}$ - log ${\displaystyle \left(z^4\right)}$

We know that log(ab) = log a + log b

⇒ log ${\displaystyle \left(x^2\right)}$ + log ${\displaystyle \left(y^3\right)}$ - log ${\displaystyle \left(z^4\right)}$

⇒ 2 log x + 3 log y - 4 log z

⇒ 2(2m - n) + 3(n - 2m) - 4(3m - 2n)

⇒ 4m - 2n + 3n - 6m - 12m + 8n

⇒ -14m + 9n

Sol:

${\displaystyle {\log }_{x}25-{\log }_{x}5=2-{\log }_x\left(\frac{1}{125}\right)}$

⇒ ${\displaystyle {\log }_x{5}^2-{\log }_{x}5=2-{\log }_x{\left(\frac{1}{5}\right)}^3}$

⇒ ${\displaystyle {\log }_x{5}^2-{\log }_{x}5=2-{\log }_x{5}^{-3}}$

⇒ ${\displaystyle 2{\log }_{x}5-{\log }_{x}5=2+3{\log }_{x}5}$

⇒ ${\displaystyle 2{\log }_{x}5-{\log }_{x}5-3{\log }_{x}5=2}$

⇒  - 2log_x5 = 2

⇒ log_x5 = -1

⇒ x^-1 = 5

⇒ ${\displaystyle \frac{1}{x}}$ = 5

⇒ x = ${\displaystyle \frac{1}{5}}$.