SELINA Solution Class 9 Chapter 8 Logarithms Exercise 8D

Question 1

If ${\displaystyle \frac{3}{2}\log a+\frac{2}{3}}$ log b - 1 = 0, find the value of a⁹.b⁴ .

Sol:

${\displaystyle \frac{3}{2}\log a+\frac{2}{3}}$ log b - 1 = 0

⇒ ${\displaystyle {\log a}^{\frac{3}{2}}+{\log b}^{\frac{2}{3}}}$ = 1

⇒ log${\displaystyle (a^{\frac{3}{2}}\times b^{\frac{2}{3}})=1}$

⇒ log${\displaystyle (a^{\frac{3}{2}}\times b^{\frac{2}{3}})=\log 10}$

⇒ ${\displaystyle (a^{\frac{3}{2}}\times b^{\frac{2}{3}})}$ = 10

⇒ ${\displaystyle {(a^{\frac{3}{2}}\times b^{\frac{2}{3}})}^6={10}^6}$

⇒ a⁹ . b⁴ = 10⁶

Question 2

${\displaystyle \frac{3}{2}\log a+\frac{2}{3}}$ log b - 1 = 0

⇒ ${\displaystyle {\log a}^{\frac{3}{2}}+{\log b}^{\frac{2}{3}}}$ = 1

⇒ log${\displaystyle (a^{\frac{3}{2}}\times b^{\frac{2}{3}})=1}$

⇒ log${\displaystyle (a^{\frac{3}{2}}\times b^{\frac{2}{3}})=\log 10}$

⇒ ${\displaystyle (a^{\frac{3}{2}}\times b^{\frac{2}{3}})}$ = 10

⇒ ${\displaystyle {(a^{\frac{3}{2}}\times b^{\frac{2}{3}})}^6={10}^6}$

⇒ a⁹ . b⁴ = 10⁶

Sol:

Given that 
x = 1 + log 2 - log 5,
y = 2 log 3 and
z = log a - log 5
Consider
x = 1 + log 2 - log 5
= log 10 + log 2 - log 5
= log( 10 x 2 ) - log 5
= log 20 - log 5
= log ${\displaystyle \frac{20}{5}}$
= log 4                   ....(1)
We have
y = 2 log3
= log 3²
= log 9                  ....(2)
Also we have
z = log a - log 5
= log${\displaystyle \frac{a}{5}}$             ....(3)
Given that x + y = 2z
∴ Subsitute the values of x, y, and z.
from (1), (2) and (3), We have
⇒ log 4 + log 9 = 2 log ${\displaystyle \frac{a}{5}}$ 

⇒ log 4 + log 9 = log${\displaystyle {\left(\frac{a}{5}\right)}^2}$

⇒ log 4 + log 9 = log${\displaystyle \left(\frac{a^2}{25}\right)}$

⇒ ${\displaystyle \log (4\times \log 9)=\log \left(\frac{a^2}{25}\right)}$

⇒ ${\displaystyle \log 36=\log \left(\frac{a^2}{25}\right)}$

⇒ ${\displaystyle \frac{a^2}{25}=36}$
⇒ a² = 36 x 25
⇒ a²  = 900
⇒ a = 30.

Question 3

If x = log 0.6; y = log 1.25 and z = log 3 - 2 log 2, find the values of :
(i) x+y- z       
(ii) 5^{x + y - z}

Sol:

Given that
x = log 0.6 , y = log 1.25, z = log 3 - 2log 2
Consider
z = log 3 - 2log 2
= log 3 - log 2²
= log 3 - log 4
= log${\displaystyle \frac{3}{4}}$
= log 0.75               ....(1)

(i) x + y - z = log 0.6 + log 1.25 - log 0.75

= log${\displaystyle \frac{0.6\times 1.25}{0.75}}$

= log${\displaystyle \left[\frac{0.75}{0.75}\right]}$

= log 1 
= 0                          ...(2)

(ii) 5^{x + y - z} = 5⁰    ...[ ∵ x + y - z = 0 from (2) ]
= 1

Question 4

If a² = log x, b³ = log y and 3a² - 2b³ = 6 log z, express y in terms of x and z .

Sol:

Given that 
a² = log x, b³ = log y and 3a² - 2b³ = 6 log z
Consider the equation,
3a² - 2b³ = 6log z
⇒ 3log x - 2log y = 6log z
⇒ logx³ - logy² = logz⁶
⇒ log ${\displaystyle \left(\frac{x^3}{y^2}\right)}$ = logz⁶

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

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

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

⇒ y = ${\displaystyle {\left(\frac{x^3}{z^6}\right)}^{\frac{1}{2}}}$

⇒ y = ${\displaystyle \left(\frac{x^{\frac{3}{2}}}{z^{\frac{6}{2}}}\right)}$

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

Question 5

If log${\displaystyle \frac{a-b}{2}=\frac{1}{2}(\log a+\log b)}$, Show that : a² + b² = 6ab. 

Sol:

log${\displaystyle \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)}$

⇒ log${\displaystyle \left(\frac{a-b}{2}\right)=\frac{1}{2}\left(\log ab\right)}$

⇒ log${\displaystyle \left(\frac{a-b}{2}\right)= {\log \left(ab\right)}^{\frac{1}{2}}}$

⇒ ${\displaystyle \left(\frac{a-b}{2}\right)={\left(ab\right)}^{\frac{1}{2}}}$

Squaring both sides we have,

${\displaystyle {\left(\frac{a-b}{2}\right)}^2=ab}$

⇒ ${\displaystyle \frac{{(a-b)}^2}{4}=ab}$

⇒ ( a - b )² = 4ab 

⇒ a² + b² - 2ab = 4ab

⇒ a² + b² = 4ab + 2ab

⇒ a² + b² = 6ab.

Question 6

log${\displaystyle \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)}$

⇒ log${\displaystyle \left(\frac{a-b}{2}\right)=\frac{1}{2}\left(\log ab\right)}$

⇒ log${\displaystyle \left(\frac{a-b}{2}\right)= {\log \left(ab\right)}^{\frac{1}{2}}}$

⇒ ${\displaystyle \left(\frac{a-b}{2}\right)={\left(ab\right)}^{\frac{1}{2}}}$

Squaring both sides we have,

${\displaystyle {\left(\frac{a-b}{2}\right)}^2=ab}$

⇒ ${\displaystyle \frac{{(a-b)}^2}{4}=ab}$

⇒ ( a - b )² = 4ab 

⇒ a² + b² - 2ab = 4ab

⇒ a² + b² = 4ab + 2ab

⇒ a² + b² = 6ab.

Sol:

Given that
a² + b² = 23ab
⇒ a² + b² + 2ab = 23ab + 2ab
⇒ a² + b² + 2ab = 25ab
⇒  ( a + b )² = 25ab
⇒ ${\displaystyle \frac{{(a+b)}^2}{25}=ab}$

⇒ ${\displaystyle {\left(\frac{a+b}{5}\right)}^2=ab}$

⇒ log${\displaystyle {\left(\frac{a+b}{5}\right)}^2}$ = ab

⇒ ${\displaystyle 2\log \left(\frac{a+b}{5}\right)}$ = log ab

⇒ log${\displaystyle \left(\frac{a+b}{5}\right)}$ = log ab

⇒ log${\displaystyle \left(\frac{a+b}{5}\right)=\frac{1}{2}(\log a+\log b)}$

Question 7

Given that
a² + b² = 23ab
⇒ a² + b² + 2ab = 23ab + 2ab
⇒ a² + b² + 2ab = 25ab
⇒  ( a + b )² = 25ab
⇒ ${\displaystyle \frac{{(a+b)}^2}{25}=ab}$

⇒ ${\displaystyle {\left(\frac{a+b}{5}\right)}^2=ab}$

⇒ log${\displaystyle {\left(\frac{a+b}{5}\right)}^2}$ = ab

⇒ ${\displaystyle 2\log \left(\frac{a+b}{5}\right)}$ = log ab

⇒ log${\displaystyle \left(\frac{a+b}{5}\right)}$ = log ab

⇒ log${\displaystyle \left(\frac{a+b}{5}\right)=\frac{1}{2}(\log a+\log b)}$

Sol:

Given that
m = log 20 and n = log 25
We also have
2log( x - 4 ) = 2m - n
⇒ 2log ( x - 4 ) = 2log 20 - log 25
⇒  log( x - 4 )² = log20² - log 25
⇒  log( x - 4 )² = log 400 - log 25
⇒  log( x - 4 )² = log ${\displaystyle \frac{400}{25}}$
⇒  ( x - 4 )² = ${\displaystyle \frac{400}{25}}$
⇒  ( x - 4 )² = 16
⇒ x - 4 = 4
⇒ x = 4 + 4
⇒ x = 8.

Question 8

Solve for x and y ; if x > 0 and y > 0 ; log xy = log ${\displaystyle \frac{x}{y}}$ + 2 log 2 = 2.

Sol:

Log xy = log${\displaystyle \left(\frac{x}{y}\right)}$ + 2log2 = 2
log xy = 2
⇒ log xy = 2log10
⇒ log xy = log 10²
⇒ log xy = log 100
∴ xy = 100                    ...(1) 
Now consider the equation
${\displaystyle \log \left(\frac{x}{y}\right)+2\log 2=2}$

⇒ ${\displaystyle \log \left(\frac{x}{y}\right)+{\log 2}^2=2\log 10}$

⇒ ${\displaystyle \log \left(\frac{x}{y}\right)+\log 4={\log 10}^2}$

⇒ ${\displaystyle \log \left(\frac{x}{y}\right)+\log 4=\log 100}$

⇒ ${\displaystyle \left(\frac{x}{y}\right)\times 4=100}$

⇒  4x = 100y
⇒ x = 25y
⇒ xy = 25y x y
⇒ xy = 25y² 
⇒ 100 = 25y²        ...[ from(1) ]

⇒ y² = ${\displaystyle \frac{100}{25}}$
⇒ y² = 4
⇒ y = 2               ....[ ∵ y > 0 ]
From (1),
xy = 100
⇒ x x 2 = 100
⇒ x = ${\displaystyle \frac{100}{2}}$
⇒ x = 50.
Thus the values of x and y are x = 50 and y = 2.

Question 9.1

Log xy = log${\displaystyle \left(\frac{x}{y}\right)}$ + 2log2 = 2
log xy = 2
⇒ log xy = 2log10
⇒ log xy = log 10²
⇒ log xy = log 100
∴ xy = 100                    ...(1) 
Now consider the equation
${\displaystyle \log \left(\frac{x}{y}\right)+2\log 2=2}$

⇒ ${\displaystyle \log \left(\frac{x}{y}\right)+{\log 2}^2=2\log 10}$

⇒ ${\displaystyle \log \left(\frac{x}{y}\right)+\log 4={\log 10}^2}$

⇒ ${\displaystyle \log \left(\frac{x}{y}\right)+\log 4=\log 100}$

⇒ ${\displaystyle \left(\frac{x}{y}\right)\times 4=100}$

⇒  4x = 100y
⇒ x = 25y
⇒ xy = 25y x y
⇒ xy = 25y² 
⇒ 100 = 25y²        ...[ from(1) ]

⇒ y² = ${\displaystyle \frac{100}{25}}$
⇒ y² = 4
⇒ y = 2               ....[ ∵ y > 0 ]
From (1),
xy = 100
⇒ x x 2 = 100
⇒ x = ${\displaystyle \frac{100}{2}}$
⇒ x = 50.
Thus the values of x and y are x = 50 and y = 2.

Sol:

log_x 625 = - 4

⇒ 625 = x^{- 4}          ...[ Removing logarithm ]

⇒ 5⁴ = ${\displaystyle {\left(\frac{1}{x}\right)}^4}$
⇒ 5 = ${\displaystyle \frac{1}{x}}$           ....[ Powers are same, bases are equal ]
⇒ x = ${\displaystyle \frac{1}{5}}$

Question 9.2

Find x, if : log_x (5x - 6) = 2

Sol:

log_x (5x - 6) = 2
⇒ 5x - 6 = x²              ...[ Removing logarithm ]
⇒ x² - 5x + 6 = 0
⇒ x² - 3x - 2x + 6 = 0
⇒ x( x - 3 ) - 2( x - 3 ) = 0
⇒ ( x - 2 )( x - 3 ) = 0
∴ x = 2, 3.

Question 9.3

Evaluate :  ${\displaystyle \frac{{\log }_5^8}{\left({\log }_{25}16\right)\times \left({\log }_{100} 10\right)}}$

Sol:

 ${\displaystyle \frac{{\log }_{5}8}{\left({\log }_{25}16\right)\times \left({\log }_{100} 10\right)}}$

⇒ ${\displaystyle \frac{\frac{{\log }_{10} 8}{\left({\log }_{10}5\right)}}{\frac{{\log }_{10}16}{{\log }_{10}25}\times \frac{{\log }_{10}10}{{\log }_{10}100}}}$

⇒ ${\displaystyle \frac{\frac{{\log }_{10}{2}^3}{\left({\log }_{10}5\right)}}{\frac{{\log }_{10}{2}^4}{{\log }_{10}{5}^2}\times \frac{{\log }_{10}10}{{\log }_{10}{10}^2}}}$ 

⇒ ${\displaystyle \frac{{\log }_{10}{2}^3}{{\log }_{10} 5}\times \frac{{\log }_{10} 5^2}{{\log }_{10}{2}^4}\times \frac{{\log }_{10} {10}^2}{{\log }_{10}10}}$ 

⇒ ${\displaystyle \frac{3{\log }_{10}2}{{\log }_{10} 5}\times \frac{2{\log }_{10} 5}{4{\log }_{10}2}\times \frac{2{\log }_{10} 10 }{{\log }_{10}10}}$ 

⇒ 3

Question 10

 ${\displaystyle \frac{{\log }_{5}8}{\left({\log }_{25}16\right)\times \left({\log }_{100} 10\right)}}$

⇒ ${\displaystyle \frac{\frac{{\log }_{10} 8}{\left({\log }_{10}5\right)}}{\frac{{\log }_{10}16}{{\log }_{10}25}\times \frac{{\log }_{10}10}{{\log }_{10}100}}}$

⇒ ${\displaystyle \frac{\frac{{\log }_{10}{2}^3}{\left({\log }_{10}5\right)}}{\frac{{\log }_{10}{2}^4}{{\log }_{10}{5}^2}\times \frac{{\log }_{10}10}{{\log }_{10}{10}^2}}}$ 

⇒ ${\displaystyle \frac{{\log }_{10}{2}^3}{{\log }_{10} 5}\times \frac{{\log }_{10} 5^2}{{\log }_{10}{2}^4}\times \frac{{\log }_{10} {10}^2}{{\log }_{10}10}}$ 

⇒ ${\displaystyle \frac{3{\log }_{10}2}{{\log }_{10} 5}\times \frac{2{\log }_{10} 5}{4{\log }_{10}2}\times \frac{2{\log }_{10} 10 }{{\log }_{10}10}}$ 

⇒ 3

Sol:

Given that
p = log 20 and q = log 25
We also have
2 log( x + 1 ) = 2p - q
⇒ 2log( x + 1 ) = 2 log 20 - log 25
⇒ log( x + 1 )² = log20² - log 25
⇒  log( x + 1 )² = log 400 - log 25
⇒ log( x + 1 )² = log${\displaystyle \frac{400}{25}}$
⇒ log( x + 1 )² = log 16
⇒ log( x + 1 )² = log 4² 
⇒ x + 1 = 4
⇒ x = 4 - 1
⇒ x = 3.

Question 11

If log₂( x + y ) = log₃( x - y ) = ${\displaystyle \frac{\log 25}{\log 0.2}}$ , find the values of x and y.

Sol:

log₂( x + y ) = ${\displaystyle \frac{\log 25}{\log 0.2}}$ 

⇒ log₂( x + y ) = log_0.2 25
⇒ log₂( x + y ) = ${\displaystyle {\log }_{\frac{2}{10}}25}$

⇒ ${\displaystyle {\log }_2(x+y)={\log }_5^{-1}{5}^2}$

⇒ ${\displaystyle {\log }_2(x+y)=-2{\log }_{5}5}$

⇒ ${\displaystyle {\log }_2(x+y)=-2}$

⇒  x + y = 2^-2                      ...[ Removing logarithm ]

⇒  x + y = ${\displaystyle \frac{1}{4}}$                    ....(1)

${\displaystyle {\log }_3(x-y)=\frac{\log 25}{\log 0.2}}$

⇒ ${\displaystyle {\log }_3(x-y)={\log }_{0.2}25}$

⇒ log₃( x - y ) = ${\displaystyle {\log }_{\frac{2}{10}}25}$

⇒ ${\displaystyle {\log }_3(x-y)={\log }_5^{-1}{5}^2}$

⇒ ${\displaystyle {\log }_3(x-y)=-2{\log }_{5}5}$

⇒ ${\displaystyle {\log }_3(x-y)=-2}$          
⇒  x - y = 3^-2                          ...[ Removing logarithm ]
⇒  x - y = ${\displaystyle \frac{1}{9}}$                   ....(2)

Solving (1) and (2), We get
x = ${\displaystyle \frac{13}{72},y=\frac{5}{72}}$

Question 12

Given : ${\displaystyle \frac{\log x}{\log y}=\frac{3}{2}}$ and log (xy) = 5; find the value of x and y.

Sol:

${\displaystyle \frac{\log x}{\log y}=\frac{3}{2}}$
⇒ 2log x = 3log y
⇒ log y = ${\displaystyle \frac{2\log x}{3}}$          ...(1)
log( xy ) = 5
⇒  log x + log y = 5
⇒  log x + ${\displaystyle \frac{2\log x}{3}}$ = 5            ....[ Substituting (1) ] 
⇒ ${\displaystyle \frac{3\log x+2\log x}{3}=5}$

⇒ ${\displaystyle \frac{5\log x}{3}=5}$
⇒ log x = 3
⇒ x = 10³ 
∴ x = 1000
Substituting x = 1000
log y = ${\displaystyle \frac{2\times 3}{3}}$  
⇒ log y = 2
⇒ y = 10²
∴ y = 100.

Question 13.1

Given log₁₀x = 2a and log₁₀y = ${\displaystyle \frac{b}{2}}$. Write 10^a in terms of x.

Sol:

log₁₀x = 2a
⇒ x = 10^2a    ...[ Removing logarithm from both sides ]
⇒ x^1/2 = 10^a
⇒ 10^a = x^1/2

Question 13.2

Given log₁₀x = 2a and log₁₀y = ${\displaystyle \frac{b}{2}}$. Write 10^{2b + 1} in terms of y.

Sol:

log₁₀y = ${\displaystyle \frac{b}{2}}$
⇒ y = 10^b/2
⇒ y^{4 =} 10^2b
⇒ 10y⁴ = 10^2b x 10 
⇒ 10^{2b + 1} = 10y⁴

Question 13.3

Given log₁₀x = 2a and log₁₀y = ${\displaystyle \frac{b}{2}.\text{If} {\log }_{10}^p=3a-2b}$, express P in terms of x and y.

SoL:

We know 10^a = x^1/2
10^b/2 = y
⇒ 10^b = y² 
${\displaystyle {\log }_{10}^p}$ = 3a - 2b
⇒  p = 10^{3a - 2b}
⇒  p = (10³)^a ÷  (10²)^b
⇒  p = ( 10a )³ ÷ ( 10^b )² 
Substituting 10^a & 10^b, We get
⇒ p = ( x^1/2 )³ ÷ ( y² )²

⇒ p = ${\displaystyle x^{\frac{3}{2}}÷ y^4}$

⇒ p = ${\displaystyle \frac{x^{\frac{3}{2}}}{y^4}}$

Question 14

Solve : log₅( x + 1 ) - 1 = 1 + log₅( x - 1 ).

Sol:

log₅( x + 1 ) - 1 = 1 + log₅( x - 1 )

⇒ log₅( x + 1 ) - log₅( x - 1 ) = 2

⇒ ${\displaystyle {\log }_5 \frac{x+1}{x-1}=2}$

⇒ ${\displaystyle \frac{x+1}{x-1}=5^2}$

⇒ ${\displaystyle \frac{x+1}{x-1}=25}$
⇒ x + 1 = 25( x - 1 )
⇒ x + 1 = 25x - 25
⇒ 25x - x = 25 + 1
⇒ 24x = 26
⇒ x = ${\displaystyle \frac{26}{24}=\frac{13}{12}}$.

Question 15

Solve for x, if : log_x49 - log_x7 + log_x ${\displaystyle \frac{1}{343}}$ + 2 = 0

Sol:

log_x49 - log_x7 + log_x ${\displaystyle \frac{1}{343}}$ = - 2

⇒ log_x${\displaystyle \frac{49}{7\times 343}}$ = - 2 

⇒ log_x${\displaystyle \frac{1}{49}}$ = - 2
⇒ - log_x 49 = - 2
⇒ log_x49 = 2
⇒ 49 = x²          ...[Removing logarithm]
∴ x = 7.

Question 16

If a² = log x , b³ = log y and ${\displaystyle \frac{a^2}{2}-\frac{b^3}{3}}$ = log c , find c in terms of x and y.

Sol:

Given a² = log x , b³ = log y

Now ${\displaystyle \frac{a^2}{2}-\frac{b^3}{3}}$ = log c

⇒ ${\displaystyle \frac{\log x}{2}-\frac{\log y}{3}=\log c}$

⇒ ${\displaystyle \frac{3\log x-2\log y}{6}=\log c}$

⇒ 3log x - 2log y = 6log c
⇒ log x³ - logy² = 6log c

⇒ ${\displaystyle \log \left(\frac{x^3}{y^2}\right)={\log c}^6}$
⇒ ${\displaystyle \frac{x^3}{y^2}=c^6}$
⇒ c = ${\displaystyle \sqrt[6]{\frac{x^3}{y^2}}}$

Question 17

Given x = log₁₀12 , y = log₄ 2 x log₁₀9 and z = log₁₀0.4 , find :
(i) x - y - z
(ii) 13^{x - y - z}

Sol:

(i) x - y - z

= log₁₀12 - log₄2 x log₁₀9 - log₁₀0.4

= log₁₀( 4 x 3 ) - log₄2 x log₁₀9 - log₁₀0.4

= log₁₀4 + log₁₀3 - log₄2 x 2log₁₀3 - log₁₀${\displaystyle \left(\frac{4}{10}\right)}$

= log₁₀4 + log₁₀3 - ${\displaystyle \frac{{\log }_{10}2}{2{\log }_{10}2}}$ x 2log₁₀3 - log₁₀4 + log₁₀10
= log₁₀4 + log₁₀3 - ${\displaystyle \frac{2{\log }_{10}3}{2}}$- log₁₀4 + 1
= 1

(ii) 13^{x - y - z} = 13¹ = 13.

Question 18

Solve for x, ${\displaystyle {\log }_x^{15\surd 5}=2-{\log }_x^{3\surd 5}}$.

Sol:

log_x15√5 = 2 - log_x3√5

⇒ log_x15√5 + log_x3√5 = 2

⇒ log_x( 15√5 x 3√5  ) = 2

⇒ log_x 225 = 2

⇒ log_x 15² = 2

⇒  2log_x 15 = 2

⇒ log_x15 = 1
⇒ x = 15.

Question 19.1

Evaluate: log_b a × log_c b × log_a c.

Sol:

log_b a x log_c b x log_a c   

⇒ ${\displaystyle \frac{{\log }_{10}a}{{\log }_{10}b}}$ x  ${\displaystyle \frac{{\log }_{10}b}{{\log }_{10}c}}$ x  ${\displaystyle \frac{{\log }_{10}c}{{\log }_{10}a}}$

⇒  1.

Question 19.2

Evaluate :  log₃8 ÷ log₉16 

SOl:

 log₃8 ÷ log₉16 

⇒ ${\displaystyle \frac{{\log }_{3}8}{{\log }_{9}16}}$

⇒ ${\displaystyle \frac{{\log }_{10}8}{{\log }_{10}3}}$ ×  ${\displaystyle \frac{{\log }_{10}9}{{\log }_{10}16}}$

⇒ ${\displaystyle \frac{3{\log }_{10}2}{{\log }_{10}3}}$ ×  ${\displaystyle \frac{2{\log }_{10}3}{4{\log }_{10}2}}$

⇒  ${\displaystyle \frac{3}{2}}$

Question 19.3

Evaluate: ${\displaystyle \frac{{\log }_{5}8}{{\log }_{25}16\times {\mathrm{Log}}_{100}10}}$ 

Sol:

${\displaystyle \frac{{\log }_{5}8}{{\log }_{25}16\times {\log }_{100}10}}$ 

= ${\displaystyle \frac{\frac{{\log }_{10}8}{{\log }_{10}5}}{\frac{{\log }_{10}16}{{\log }_{10}25}\times \frac{{\log }_{10}10}{{\log }_{10}100}}}$

= ${\displaystyle \frac{\frac{{\log }_{10}{2}^3}{{\log }_{10}5}}{\frac{{\log }_{10}{2}^4}{{\log }_{10}{5}^2}\times \frac{{\log }_{10}10}{{\log }_{10}{10}^2}}}$

= ${\displaystyle \frac{{\log }_{10}{2}^3}{{\log }_{10}5}\times \frac{{\log }_{10}{5}^2}{{\log }_{10}{2}^4}\times \frac{{\log }_{10}{10}^2}{{\log }_{10}10}}$

= ${\displaystyle \frac{3{\log }_{10}2}{{\log }_{10}5}\times \frac{2{\log }_{10}5}{4{\log }_{10}2}\times \frac{2{\log }_{10}10}{{\log }_{10}10}}$

= 3

Question 20

Show that : log_a m ÷ log_ab m + 1 + log _ab 

SOl:

log_a m ÷ log_ab m = =${\displaystyle \frac{{\log }_{a}m}{{\log }_{ab}m}}$  

=${\displaystyle \frac{{\log }_{m}ab}{{\log }_{m}a}[Q{\log }_{b}a=\frac{1}{{\log }_{a}b}]}$

₌  log_a ab  ${\displaystyle [Q\frac{{\log }_{\times }a}{{\log }_{\times }b}={\log }_{b}a]}$

 = log_a a + log_a b   

 = 1 + log_a b 

Question 21

If log_√27x = 2 ${\displaystyle \frac{2}{3}}$ , find x.

SOl:

log_√27x = 2 ${\displaystyle \frac{2}{3}}$

∴ log_√27x =  ${\displaystyle \frac{8}{3}}$

∴ x = ${\displaystyle {\left(\sqrt{27}\right)}^{\frac{8}{3}}}$  ...[ ∵ log_a x = b ⇒ x = a^b ]     

∴ x = ${\displaystyle {\left({27}^{\frac{1}{2}}\right)}^{\frac{8}{3}}}$ 

∴ x = ${\displaystyle {\left(3^{\frac{3}{2}}\right)}^{\frac{8}{3}}}$

∴ x = ${\displaystyle 3^{\left(\frac{3}{2}\right)\times \left(\frac{8}{3}\right)}}$

∴ x = 3⁴ 

∴ x = 81

Question 22

Evaluate :${\displaystyle \frac{1}{{\log }_{a}bc+1}+\frac{1}{{\log }_{b}ca+1}+\frac{1}{{\log }_{c}ab+1}}$

Sol:

⇒ ${\displaystyle \frac{1}{{\log }_{a}bc+1}+\frac{1}{{\log }_{b}ca+1}+\frac{1}{{\log }_{c}ab+1}}$

⇒${\displaystyle \frac{1}{{\log }_{a}bc+{\log }_{a}a}+\frac{1}{{\log }_{b}ca+{\log }_{b}b}+\frac{1}{{\log }_{c}ab+{\log }_{c}c}}$

⇒ ${\displaystyle \frac{1}{{\log }_{a}abc}+\frac{1}{{\log }_{b}abc}+\frac{1}{{\log }_{c}abc}}$  ...[∵ log_a b + log_a c = log_a bc ]

⇒ ${\displaystyle \frac{1}{\frac{\log abc}{\log a}}}$ + ${\displaystyle \frac{1}{\frac{\log abc}{\log b}}}$ + ${\displaystyle \frac{1}{\frac{\log abc}{\log c}}}$

⇒ ${\displaystyle \frac{\log a+\log b+\log c}{\log abc}}$

⇒ ${\displaystyle \frac{\log abc}{\log abc}}$  ...∵[ log_a b + log_a c = log_a bc ] 

⇒ 1