SELINA Solution Class 9 Chapter 8 Logarithms Exercise 8A

Question 1

Express the following in logarithmic form : 5³ = 125

Sol:

5³ = 125
⇒ log₅125 = 3     ...[ a^b = c ⇒  log_ac = b ]

Question 1.2

Express the following in logarithmic form :
3^{-2 =} ${\displaystyle \frac{1}{9}}$

Sol:

3^{-2 =} ${\displaystyle \frac{1}{9}}$
⇒ log₃${\displaystyle \frac{1}{9}}$ = - 2     ....[ a^b = c ⇒  log_ac = b ]

Question 1.3

Express the following in logarithmic form :
10^-3 = 0.001

Sol:

10^-3 = 0.001
⇒ log₁₀0.001 = - 3   ....[ a^b = c ⇒  log_ac = b ]

Question 1.4

Express the following in logarithmic form : ${\displaystyle {\left(81\right)}^{\frac{3}{4}}=27}$ 

Sol:

${\displaystyle {\left(81\right)}^{\frac{3}{4}}=27}$ 

⇒ log₈₁27 = ${\displaystyle \frac{3}{4}}$   ....[ By definition of logarithm, a^b = c ⇒ log_ac = b ]

Question 2.1

Express the following in exponential form : log₈0.125 = -1

SoL:

log₈0.125 = -1
⇒ 8^-1 = 0.125      ....[ log_ac = b ⇒ a^b = c ]

Question 2.2

Express the following in exponential form : 
log₁₀0.01 = - 2

Sol;

log₁₀0.01 = - 2
⇒ 10^-2 = 0.01  .....[ log_ac = b ⇒ a^b = c ]

Question 2.3

Express the following in exponential form : log_aA = x

Sol:

log_aA = x
⇒ a^x = A .....[ log_ac = b ⇒ a^b = c  ]

Question 2.4

Express the following in exponential form : log₁₀1 = 0

Sol:

log₁₀1 = 0
⇒ 10⁰ = 1    .....[ log_ac = b ⇒ a^b = c ]

Question 3

Solve for x : log₁₀ x = -2.

Sol:

log₁₀ x = -2
⇒ 10^-2 = x          ....[ log_ac = b ⇒ a^b = c ] 
⇒ x = 10^-2

⇒  x = ${\displaystyle \frac{1}{{10}^2}}$

⇒ x = ${\displaystyle \frac{1}{100}}$

⇒ x = 0.01

Question 4.1

Find the logarithm of : 100 to the base 10

Sol:

Let log₁₀100 = x 
∴ 10^x = 100
⇒ 10^x = 10 x 10
⇒  10^x = 10² 
⇒ x = 2          .....[ If a^m = aⁿ ; then m = n ]
∴ log₁₀100 = 2

Question 4.2

Find the logarithm of : 0.1 to the base 10

SoL:

Let log₁₀0.1 = x
∴ 10^x = 0.1
⇒ 10^x = ${\displaystyle \frac{1}{10}}$
⇒  10^x = 10^-1
⇒  x = - 1       ......[ If a^m = aⁿ ; then m = n ]
∴ log₁₀0.1 = - 1 

Question 4.3

Find the logarithm of : 0.001 to the base 10

Sol:

Let log₁₀0.001 = x 
∴ 10^x = 0.001
⇒ 10^x = ${\displaystyle \frac{1}{1000}}$

⇒ 10^x = ${\displaystyle \frac{1}{{10}^3}}$
⇒  10^x = 10^-3
⇒ x = - 3        ......[ If a^m = aⁿ ; then m = n ]

∴ log₁₀0.001 = - 3

Question 4.4

Find the logarithm of : 32 to the base 4

Sol:

Let log₄32 = x
∴ 4^x = 32
⇒ (2²)^x = 2 x 2 x 2 x 2 x 2
⇒ 2^2x = 2⁵
⇒ 2x = 5                  .....[ If a^m = aⁿ ; then m = n ]
⇒ x = ${\displaystyle \frac{5}{2}}$
∴ log₄32 = ${\displaystyle \frac{5}{2}}$

Question 4.5

Find the logarithm of : 0.125 to the base 2

Sol:

Let log₂0.125 = x
∴ 2^x = 0.125
⇒ 2^x = ${\displaystyle \frac{125}{1000}}$
⇒ 2^x = ${\displaystyle \frac{1}{8}}$
⇒ 2^x = 8^-1
⇒ 2^x = ( 2 x 2 x 2 )^-1
⇒ 2^x = ( 2³ )^-1
⇒ 2^x = 2^-3
⇒ x = - 3     .....[ If a^m = aⁿ ; then m = n ]
∴ log₂0.125 = -3

Question 4.6

Find the logarithm of : ${\displaystyle \frac{1}{16}}$ to the base 4

Sol:

Let log₄${\displaystyle \frac{1}{16}}$ = x

∴ 4^x = ${\displaystyle \frac{1}{16}}$

⇒ 4^x = ${\displaystyle \frac{1}{4\times 4}}$

⇒ 4^x = ( 4 x 4 )^-1

⇒ 4^x = ( 4² )^-1

⇒ 4^x = 4^-2

⇒ x = - 2      ....[ If a^m = aⁿ ; then m = n ]

∴ log₄${\displaystyle \frac{1}{16}}$ = - 2

Question 4.7

Find the logarithm of : 27 to the base 9

SoL:

Let log₉27 = x
∴ 9^x = 27
⇒ ( 3 x 3 )^x = 3 x 3 x 3
⇒ (3²)^x = 3³
⇒ (3²)^x = 3³
⇒ 2x = 3         .....[ If a^m = aⁿ ; then m = n ]
⇒ x = ${\displaystyle \frac{3}{2}}$
∴ log₉27 = ${\displaystyle \frac{3}{2}}$

Question 4.8

Find the logarithm of : ${\displaystyle \frac{1}{81}}$ to the base 27

Sol:

Let log₂₇${\displaystyle \frac{1}{81}}$ = x

∴ 27^x = ${\displaystyle \frac{1}{81}}$

⇒ ( 3 x 3 x 3 )^x = ${\displaystyle \frac{1}{3\times 3\times 3\times 3}}$

⇒  (3³)^x = ${\displaystyle \frac{1}{3^4}}$

⇒  (3³)^x = (3⁴)^-1

⇒  3^3x = 3^-4

⇒ 3x = - 4           .....[ If a^m = aⁿ ; then m = n ] 

⇒ x = ${\displaystyle -\frac{4}{3}}$

∴ log₂₇${\displaystyle \frac{1}{81}=-\frac{4}{3}}$

Find the logarithm of : ${\displaystyle \frac{1}{81}}$ to the base 27

Question 5.1

State, true or false : If log₁₀ x = a, then 10^x = a.

Sol:

Consider the equation
log₁₀x = a
⇒ 10^a = x
Thus the statement, 10^x = a is false.

Question 5.2

State, true or false : If x^y = z, then y = log_zx .

Sol:

Consider the equation
x^y = z
⇒ log_xz = y
Thus the statement, log_zx = y is false.

Question 5.3

State, true or false : log₂ 8 = 3 and log₈ = 2 = ${\displaystyle \frac{1}{3}}$

Sol:

Consider the equation
log₂8 = 3
⇒ 2³ = 8                                ....(1)
Now consider the equation
log₈2 = ${\displaystyle \frac{1}{3}}$
⇒ ${\displaystyle 8^{\frac{1}{3}}=2}$

⇒ ${\displaystyle {\left(2^3\right)}^{\frac{1}{3}}=2}$                ...(2)

Both the equations (1) and (2) are correct.
Thus the given statements, log₂8 = 3 and log₈2 = ${\displaystyle \frac{1}{3}}$ are true.

Question 6.1

Find x, if : log₃ x = 0

SoL:

Consider the equation
log₃x = 0
⇒ 3⁰ = x
⇒ 1 = x or x = 1

Question 6.2

Find x, if : log_x 2 = - 1.

SoL:

Consider the equation
log_x 2 = - 1
⇒ x^-1 = 2

⇒ ${\displaystyle \frac{1}{x}}$ = 2

⇒ x = ${\displaystyle \frac{1}{2}}$

Question 6.3

Find x, if : log₉243 = x

Sol:

Consider the equation
log₉243 = x
⇒ 9^x = 243
⇒ (3²)^x = 3⁵

⇒ 3^2x = 3⁵

⇒ 2x = 5

⇒ x = ${\displaystyle \frac{5}{2}}$

⇒ x = ${\displaystyle 2\frac{1}{2}}$

Question 6.4

Find x, if : log₅ (x - 7) = 1

Sol:

Consider the equation
log₅ (x - 7) = 1
⇒ 5¹ = x - 7
⇒  5 = x - 7
⇒ x = 5 + 7
⇒ x = 12 

Question 6.5

Find x, if : log₄32 = x - 4

Sol:

Consider the equation
log₄32 = x - 4
⇒ 4^{x - 4} = 32
⇒  (2²)^{x - 4} = 2⁵
⇒  2^{2( x - 4 )} = 2⁵
⇒ 2x - 8 = 5 
⇒ 2x = 5 + 8 
⇒ 2x = 13

⇒ x = ${\displaystyle \frac{13}{2}}$

⇒ x = ${\displaystyle 6\frac{1}{2}}$

Question 6.6

Find x, if : log₇ (2x² - 1) = 2

SoL:

Consider the equation
log₇( 2x² - 1 ) = 2
⇒ 7² = 2x² - 1
⇒ 7 x 7 = 2x² - 1
⇒ 2x² - 1 - 49 = 0
⇒ 2x² - 50 = 0
⇒ 2x² = 50
⇒ x² = ${\displaystyle \frac{50}{2}}$
⇒ x² = 25
⇒ x = ${\displaystyle \pm \sqrt{25}}$
⇒ x = 5               .....[ neglecting the negative value ]

Question 7.1

Evaluate : log₁₀ 0.01

Sol:

Let log₁₀ 0.01 = x
⇒ 10^x = 0.01
⇒ 10^x = ${\displaystyle \frac{1}{100}}$
⇒ 10^x = ${\displaystyle \frac{1}{10\times 10}}$
⇒ 10^x = ${\displaystyle \frac{1}{{10}^2}}$
⇒ 10^x = 10^-2
⇒ x = - 2
Thus, log₁₀ 0.01 = - 2

Question 7.3

Evaluate : log₅ 1

Sol:

Let log₅ 1 = x
⇒ 5^x = 1
⇒ 5^x = 5⁰
⇒ x = 0
Thus, log₅ 1 = 0

Question 7.4

Evaluate : log₅ 125

Sol:

Let log₅ 125 = x
⇒ 5^x = 125
⇒ 5^x = 5 x 5 x 5
⇒ 5^x = 5³
⇒ x = 3
Thus, log₅125 = 3

Question 7.5

Evaluate : log₁₆ 8

Sol:

Let log₁₆ 8 = x
⇒ 16^x = 8
⇒  ( 2 x 2 x 2 x 2 )^x = 2 x 2 x 2
⇒ ( 2⁴ )^x = 2³
⇒ 2^4x = 2³
⇒ 4x = 3 
⇒ x = ${\displaystyle \frac{3}{4}}$
Thus, log₁₆ 8 = ${\displaystyle \frac{3}{4}}$

Question 7.6

Evaluate : log_0.5 16

Sol:

Let log_0.5 16 = x
⇒ 0.5^x = 16
⇒ ${\displaystyle {\left(\frac{5}{10}\right)}^x=2\times 2\times 2\times 2}$

⇒ ${\displaystyle {\left(\frac{1}{2}\right)}^x=2^4}$

⇒ ${\displaystyle \frac{1}{2^x}=2^4}$

⇒ 2^{- x} = 2⁴
⇒ - x = 4
⇒ x = - 4
Thus, log_0.516 = - 4

Question 8

If log_a m = n, express a^{n - 1} in terms of a and m.

Sol:

log_a m = n

⇒ aⁿ  = m

⇒ ${\displaystyle \frac{a^n}{a}=\frac{m}{a}}$

⇒ ${\displaystyle a^{n-1}=\frac{m}{a}}$

Question 9.1

Given log₂ x = m. Express 2^{m - 3}  in terms of x.

Sol:

log₂x  = m

⇒ 2^m = x 
Consider 2^m  = x

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

⇒ ${\displaystyle 2^{m-3}=\frac{x}{8}}$

Question 9.2

Given log₅ y = n. Express 5^{3n + 2} in terms of y.

Sol:

log₅ y = n
5ⁿ = y
Consider 5ⁿ = y
⇒ ${\displaystyle {\left(5^n\right)}^3=y^3}$
⇒ ${\displaystyle 5^{3n}=y^3}$ 
⇒ ${\displaystyle 5^{3n}\times 5^2=y^3\times 5^2}$
⇒${\displaystyle 5^{3n+2}=25y^3}$

Question 10

If log₂x = a  and log₃ y = a, write 72^a in terms of x and y.

SoL:

Given that :
log₂x = a and log₃y = a 
⇒ 2^a = x and 3^a = y       ....[ Q log_am = n ⇒  aⁿ = m ]
Now prime factorization of 72 is 
72 = 2 x 2 x 2 x 3 x 3
Hence,
(72)^a = (2 x 2  x 2 x 3 x 3)^a
        = (2³ x 3²)^a 
       = 2^3a x 3^2a 
       =  (2^a)³ x (3^a)²        ....[ as 2^a = x , 3^a = y ]
       =  x³y²

Question 11

Solve for x : log( x - 1 ) + log (x + 1 ) = log₂1.

Sol:

log ( x - 1) + log( x + 1) = log₂1
⇒ log ( x - 1 ) + log(x + 1 )= 0
⇒ log [ ( x -1 ) ( x + 1) ] = 0
⇒ ( x - 1 ) ( x + 1) = 1.... ( Since log 1 = 0 )
⇒ x² - 1 = 1
⇒ x² = 2
⇒ x = ${\displaystyle \pm \sqrt{2}}$
 ${\displaystyle -\sqrt{2}}$ cannot be possible, since log of a negative number is not defined.
So, x = ${\displaystyle \sqrt{2}}$.

Question 12

If log (x² - 21) = 2, show that x = ± 11.

Sol:

log (x²  - 21 ) = 2
⇒ x²  - 21 = 10^2
⇒  x² - 21 = 100
⇒ x² = 121
⇒ x = ${\displaystyle \pm 11}$