Question 1: Simplify each of the following:

(i) \log x^5 - \log  x^4 = \log \frac{x^5}{x^4} = \log  x

(ii) \log x^5 \div \log x^4 = \frac{\log x^5}{\log x^4} = \frac{5 \log x}{4 \log x} = \frac{5}{4} 

(iii) \frac{\log 27}{\log 9 } = \frac{\log 3^3}{\log  3^2} = \frac{3}{2} 

(iv) \log_6 72 - \log_6 2 = \log_6 \frac{72}{2} = \log_6 36 = \log_6 6^2 = 2

(v) \frac{\log 27 - \log 9}{\log 81} = \frac{3 \log 3 - 2 \log 3}{4 \log 3} = \frac{1}{4} 

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Question 2: Prove the following

(i) \log 108 = 3 \log  3 + 2 \log  2

LHS = \log 108 =  \log (27 \times 4) =  \log 27 +  \log 4 = 3  \log 3 + 2  \log  2 = RHS

(ii) \log  75 = \log  3 + 2 \log  5

LHS =  \log 75 =  \log (25 \times 3) =  \log 25 +  \log 3 = 2  \log 5 +  \log  3 = RHS

(iii) \log 72 = 2 \log 3 + 3 \log  2

LHS =  \log  (8 \times 9) =  \log 8 +  \log 9 = 3  \log 2 + 2  \log 3 = RHS

(iv) 2 \log  5 + \log  8 - \frac{1}{2} \log 4 = 2

LHS =  \log 25 +  \log 8 -  \log 2 =  \log \frac{25 \times 8}{2} =  \log 100 =  \log 10^2 = 2 = RHS

(v) \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} = 0

LHS = \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} = \log \Big(  \frac{4}{7} \times  \frac{33}{18} \times  \frac{21}{22} \Big) = \log 1 = 0

(vi) \log  2 + 16 \log \frac{16}{15} + 12  \log \frac{25}{24} +7  \log \frac{81}{80} = 1

LHS = \log  2 + \log ( \frac{16}{15} )^{16} + \log ( \frac{25}{24} )^{12} + \log  ( \frac{81}{80} )^7

= \log \Big( 2 \times ( \frac{16}{15} )^{16} \times ( \frac{25}{24} )^{12} \times ( \frac{81}{80} )^7 \Big) = \log 10 = 1

(vii) \log 2 + 2 \log 5 - \log  3 - 2 \log 7 = \log  \frac{50}{147} 

LHS = \log 2 + 2 \log 5 - \log  3 - 2 \log 7 = \log \Big( \frac{2 \times 5^2}{3 \times 7^2} \Big) = \log \frac{50}{147}  

(viii) 2\log \frac{11}{13}   + \log \frac{130}{77}   - \log \frac{55}{91}   = \log 2

LHS = 2\log \frac{11}{13}   + \log \frac{130}{77}   -  \log \frac{55}{77} = \log \Big( ( \frac{11}{13} )^2 \times  \frac{130}{77} \times  \frac{77}{91} \Big) = \log \frac{110}{55} = \log 2

(ix)  \log \frac{75}{16} + \log \frac{32}{243} - 2 \log \frac{5}{9} = \log  2

LHS = \log \frac{75}{16} + \log \frac{32}{243} - 2 \log \frac{5}{9} = \log ( \frac{75}{16} \times  \frac{32}{243} \times ( \frac{9}{5})^2) = \log 2

(x) 7 \log \frac{10}{9} + 3 \log \frac{81}{80} - 2 \log \frac{25}{24} = \log 2

LHS = 7 \log \frac{10}{9} + 3 \log \frac{81}{80} - 2 \log \frac{25}{24} = \log ( ( \frac{10}{9} )^7 \times ( \frac{81}{80} )^3 \times ( \frac{24}{25} )^2 ) = \log 2

(xi) \log (1^{1/5} + 32^{1/5} + 243^{1/5}) = \frac{1}{5} (\log  1 + \log  32 + \log  243 )

LHS = \log (1^{1/5} + 32^{1/5} + 243^{1/5}) = \log (1 + 2 + 3 ) = \log 6

RHS =  \frac{1}{5} (\log  1 + \log  32 + \log  243 ) = \log 1^{1/5} + \log 32^{1/5} + \log 243^{1/5}

= \log 1 + \log 2 + \log 3  = \log (1 \times 2 \times 3 ) = \log 6

Hence LHS = RHS

(xii) 7 \log_2 \frac{16}{15} + 5 \log_2 \frac{25}{24} + 3 \log_2 \frac{81}{80} = 1

LHS = 7 \log_2 \frac{16}{15} + 5 \log_2 \frac{25}{24} + 3 \log_2 \frac{81}{80}

= \log_2 \Big( ( \frac{16}{15} )^7 \times ( \frac{25}{24} )^5 \times ( \frac{81}{80} )^3 \Big) = \log_2 2 = 1

(xiii) \log \Big( \frac{x^2}{yz} \Big) + \log \Big( \frac{y^2}{zx} \Big) + \log \Big( \frac{z^2}{xy} \Big) = 0

LHS = \log \Big( \frac{x^2}{yz} \Big) + \log \Big( \frac{y^2}{zx} \Big) + \log \Big( \frac{z^2}{xy} \Big)

= \log \Big(  \frac{x^2}{yz} \times  \frac{y^2}{zx} \times  \frac{z^2}{xy} \Big) = \log \frac{x^2.y^2.z^2}{x^2.y^2.z^2} = \log 1 = 0

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Question 3: Express each of the following as the logarithm of a single number.

(i) \log 3 + 2  = \log 3 + \log 100 = \log 300

(ii) 2 \log 3 + 3 = \log 3^2 + \log 1000 = \log 9000

(iii) 1 + \frac{1}{3}  \log 27 = \log 10 + \log 27^{1/3} = \log 10 + \log 3 = \log 30

(iv) \frac{1}{2} \log 4 + \frac{1}{4}  \log 81 + 3 \log 2 - \log  6+2

= \log 2 + \log 3 + \log 8 - \log 6 + \log 100 = \log \frac{2 \times 3 \times 8 \times 100}{6} = \log 800

(v) \frac{1}{2}  \log  9 + 2 \log  3 - \log 6 + \log 2 - 2  = \log 3 + \log 9 - \log 6  + \log 2 - \log 100

= \log \frac{3 \times 9 \times 2}{6 \times 100} = \log \frac{9}{100}

(vi) 2 + \frac{1}{2} \log_{10} 9 - 2 \log_{10} 5 = \log_{10} 100 + \log_{10} 3 - \log_{10} 25 = \log_{10} \frac{100 \times 3}{25} = \log 12

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Question 4: Evaluate the following:

(i) 2 \log 5 + \frac{1}{3} \log 64 = \log 25 + \log 4 = \log 100 = \log 10^2 = 2

(ii) \log 21 + \log 4 + 2 \log 5 - \log 3 - \log 7 = \log \frac{21 \times 4 \times 25}{3 \times 7} = \log 100 = 2

(iii) \log 15 + 2 \log 0.5 + 3 \log 2 - \log 3 - \log 5 = \log \frac{15 \times 0.5 \times 0.5 \times 8}{3 \times 5} = \log 2

(iv) \log_3 5 \times \log_{25} 27 = \log_3 5 \times \log_{5^2} 3^3 = \log_3 5 \times \frac{3}{2} \log_5 3 = \frac{3}{2}

(v) \log_4 \{\log_{\sqrt{2}} (\log_3 81) \} = \log_4 \{ \log_{\sqrt{2}} (\log_3 3^4 ) \}

= \log_4 (\log_{\sqrt{2}} 4 ) = \log_4 (\log_{\sqrt{2}} (\sqrt{2})^4 = \log_4 4 = 1

(vi) \log_3 \sqrt{\frac{2}{3}} - \log_3 (\log_3 9))  = \log_3 \sqrt{6} + \log_3 \sqrt{\frac{2}{3}} - \log_3 2

= \log_3 \Big( \frac{\sqrt{6} \times \sqrt{2}}{2 \times \sqrt{3}} \Big) = \log_3 1 = 0  

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Question 5: Prove that:

(i) \log_{10} 4 \div \log_{10} 2 = \log_3 9

LHS = 2 \log_{10} 2 \div \log_{10} 2 = 2

RHS = \log_3 3^2 = 2 \log_3 3 = 2 . Hence proved.

(ii) \log_{10} 25 + \log_{10} 4 = \log_5 25

LHS = 2 \log_{10} 5 + 2 \log_{10} 2 = 2(\log_{10} 5+ \log_{10} 2) = 2 \log_{10} 10 = 2

RHS = \log_5 25 = \log_5 5^2 = 2 \log_5 5 = 2 . Hence proved.

Question 6: If x = 100^a, y = 10000^b and z = 10^c , express \log \Big( \frac{10\sqrt{7}}{x^2z^3} \Big) in terms of a, b \ and \  c .

Answer:

x = 100^a \Rightarrow \log x = 2a

y = 10^{4b} \Rightarrow \log y = 4b

z = 10^c \Rightarrow \log z = c

\log \Big( \frac{10\sqrt{7}}{x^2z^3} \Big)

= \log 10 + \log \sqrt{y} - \log x^2 - \log z^3

= 1 + \frac{1}{2} \log y - 2 \log x - 3 \log z

= 1 + \frac{1}{2} . 4b - 2. 2a - 3 c

= 1 +2b-4a-3c

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Question 7: If 2 \log x and 2 \log y -2 = 0 , prove that x^2y^3 = 100

Answer:

2 \log x and 2 \log y -2 = 0

\Rightarrow \log x^2 + \log y^3 - \log 100 = \log 1

\Rightarrow  \log \Big( \frac{x^2y^3}{100} \Big) = \log 1

\Rightarrow  \frac{x^2y^3}{100} = 1 

\Rightarrow x^2y^3 = 100

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Question 8: If \log \Big( \frac{a+b}{3} \Big) = \frac{1}{2} (\log a + \log b) , prove that a^2 + b^2 = 7ab

Answer:

\log \Big( \frac{a+b}{3} \Big) = \frac{1}{2} (\log a + \log b)

\Rightarrow \log \Big( \frac{a+b}{3} \Big) = \frac{1}{2} (\log ab)

\Rightarrow \log \Big( \frac{a+b}{3} \Big) =  (\log (ab)^{\frac{1}{2}})

\Rightarrow \frac{a+b}{3} = (ab)^{\frac{1}{2}}

\Rightarrow a^2 + b^2 +2ab = 9 ab 

\Rightarrow a^2 + b^2 = 7 ab

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Question 9: If a^{2x-3} b^{2x} = a^{6-x} b^{5x} , prove that 3 \log a = x \log \frac{a}{b} 

Answer:

a^{2x-3} b^{2x} = a^{6-x} b^{5x}

\Rightarrow \frac{a^{2x-3}}{a^{6-x}} = \frac{b^{5x}}{b^{2x}}

\Rightarrow a^{2x-3-6+x} = b^{5x-3x}

\Rightarrow a^{3x-9} = b^{3x}

\Rightarrow (a^{x-3})^3 = (b^x)^3

\Rightarrow a^{x-3} = b^x

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Question 10: If \log \Big( \frac{a+b}{2} \Big) = \frac{1}{2} (\log a + \log b) , prove that a = b .

Answer:

\log \Big( \frac{a+b}{2} \Big) = \frac{1}{2} (\log a + \log b)

\Rightarrow \log \Big( \frac{a+b}{2} \Big) = (\log a^{\frac{1}{2}} + \log b^{\frac{1}{2}})

\Rightarrow \log \Big( \frac{a+b}{2} \Big) = (\log (ab)^{\frac{1}{2}})

\Rightarrow \frac{a+b}{3} = (ab)^{\frac{1}{2}}

\Rightarrow \frac{a^2 + b^2 + 2ab}{4} = ab

\Rightarrow a^2 + b^2 -2ab = 0

\Rightarrow (a-b)^2 = 0

\Rightarrow a-b = 0 \ or \  a = b

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Question 11: If \log_{10} a + \frac{1}{2} \log_{10} b = 1 , prove that ba^2 = 100

Answer:

\log_{10} a + \frac{1}{2} \log_{10} b = 1

\Rightarrow \log_{10} a + \frac{1}{2} \log_{10} b = \log_{10} 10

\Rightarrow \log_{10} ab^{1/2} = \log_{10} 10

\Rightarrow ab^{\frac{1}{2}} = 10 

\Rightarrow a^2b = 100

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Question 12: If \log \frac{a-b}{2} = \frac{1}{2} (\log a + \log b) , prove that a^2 + b^2 = 6ab

Answer:

If \log \frac{a-b}{2} = \frac{1}{2} (\log a + \log b)

\Rightarrow \log \frac{a-b}{2} = (\log a^{\frac{1}{2}} + \log b^{\frac{1}{2}})

\Rightarrow \log \frac{a-b}{2} = (\log (ab)^{\frac{1}{2}})

\Rightarrow \frac{a-b}{2} = (ab)^{\frac{1}{2}}

\Rightarrow a^2 + b^2 -2ab = 4ab

\Rightarrow a^2 + b^2 = 6ab

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Question 13: If a^2 + b^2 = 23 ab , prove that \log \Big( \frac{a+b}{5} \Big) = \frac{1}{2} (\log a + \log b)

Answer:

a^2 + b^2 = 23 ab

\Rightarrow a^2+b^2 + 2ab = 25 ab

\Rightarrow (a+b)^2 = 25ab

\Rightarrow \log (a+b)^2 = \log 5^2ab

\Rightarrow 2 \log (a+b) = 2 \log 5 + \log a + \log b

\Rightarrow 2 \log (a+b) - 2\log 5 = \log a + \log b

\Rightarrow \log \frac{a+b}{5} = \frac{1}{2} (\log a + \log b)

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Question 14: If y = \log_{10} x , find the following in terms of y

(i) y = \log_{10} x  \Rightarrow 10^y = x

(ii) \log_{10} \sqrt[3]{x^2} = \log_{10} (x^2)^{1/3} = \frac{2}{3} \log_{10} x = \frac{2}{3} y 

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Question 15: If x = \log \frac{2}{3} , y = \log \frac{3}{5} and z =2 \log \sqrt{\frac{5}{2}} , find the value of x + y + z and 5^{x+y+z}

Answer:

Given x = \log \frac{2}{3} , y = \log \frac{3}{5} and z = 2\log \sqrt{\frac{5}{2}}

Therefore x + y + z = \log \frac{2}{3} + \log \frac{3}{5} + 2 \log \sqrt{\frac{5}{2}}  = \log \Big(   \frac{2}{3} \times \frac{3}{5} \times  \frac{5}{2} \Big) = \log 1 = 0

Therefore 5^{x+y+z} = 5^0 = 1

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Question 16: If x = \log \frac{3}{5} , y = \log \frac{5}{4} and z = 2\log \frac{\sqrt{3}}{2} , find the value of x + y - z and 7^{x+y-z}

Answer:

Given x = \log \frac{3}{5} , y = \log \frac{5}{4} and z = 2\log \frac{\sqrt{3}}{2}

x + y - z = \log \frac{3}{5}  + \log \frac{5}{4}  - 2 \log \frac{\sqrt{3}}{2} = \log \Big(  \frac{3}{5} \times  \frac{5}{4} \times  \frac{4}{3}   \Big) = \log 1 = 0

Therefore 7^{x+y-z} = 7^0 = 1

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Question 17: If x^2 = \log_{10} a, y^3 = \log_{10} b and \frac{x^2}{2} - \frac{y^3}{3} = \log_{10} c , express c in terms of a and b .

Answer:

Given \log_{10} c = \frac{x^2}{2} - \frac{y^3}{3}

\log_{10} c = \frac{1}{2}  \log_{10} a - \frac{1}{3}  \log_{10}  b

= \log_{10} a^{\frac{1}{2}} - \log_{10} b^{\frac{1}{3}}

= \log_{10} \frac{a^{\frac{1}{2}}}{b^{\frac{1}{3}}} 

c = \frac{a^{\frac{1}{2}}}{b^{\frac{1}{3}}}

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Question 18: If 2 \log_{10} a + \log_{10} b = 2 , express b in terms of a

Answer:

2 \log_{10} a + \log_{10} b = 2

2 \log_{10} a + \log_{10} b = \log_{10} 100

\log_{10} b = \log_{10} 100 -\log_{10} a^2

\log_{10} b = \log_{10} \frac{100}{a^2} 

\Rightarrow b =  \frac{100}{a^2} 

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Question 19: If  \log x = a + b and \log y = a - b , express \log \Big( \frac{10x}{y^2} \Big) in terms of a and b .

Answer:

Given \log x = a + b and \log y = a - b

\log \Big( \frac{10x}{y^2} \Big) = \log 10x - \log y^2

= \log 10 + \log x  - 2 \log y

= 1 + a  + b -2(a-b)

= 3b-a+1

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Question 20: If \log x = a + b and \log y = a - b , express \log x^2y in terms of a and b .

Answer:

Given: \log x = a + b and \log y = a - b

\log x^2y = \log x^2 + \log y

= 2 \log x + \log y

= 2 (a+b) + a-b

= 3a+b

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Question 21: Solve the following:

(i) \log (x+1) - \log  (x-1) = 1 \Rightarrow \log \frac{x+1}{x-1} = \log 10

\Rightarrow \frac{x+1}{x-1} = 10 \Rightarrow x+1 = 10x-10 \Rightarrow x = \frac{11}{9}

(ii) \log (2x+1) - \log (2x-1) = 1 \Rightarrow \log \frac{2x+1}{2x-1} = \log 10

\Rightarrow \frac{2x+1}{2x-1} = 10 \Rightarrow 2x+1 = 20x - 10 \Rightarrow x = \frac{11}{18} 

(iii) 3^{\log x} - 2^{\log x} = 2^{\log x+1}-3^{\log x-1}

\Rightarrow 3^{\log x} + \frac{3^{\log x}}{3} = 2.2^{\log x} +  2^{\log x}

\Rightarrow \frac{4}{3} \times 3^{\log x} = 3 \times 2^{\log x}

\Rightarrow \frac{3^{\log x}}{2^{\log x}} = \frac{9}{4} 

\Rightarrow \Big( \frac{3}{2} \Big)^{\log x} = \Big( \frac{3}{2} \Big)^2

\Rightarrow \log x = 2 \Rightarrow \log x = \log 100 \Rightarrow x = 100

(iv) \log_2 x + \log_4 x + \log_{64} x = 5

 \Rightarrow \log_2 x + \log_{2^2} x^1 + \log_{2^6} x^1 = 5

 \Rightarrow \log_2 x + \frac{1}{2} \log_2 x + \frac{1}{6} \log_2 x = 5

 \Rightarrow (1+ \frac{1}{2} + \frac{1}{6} ) \log_2 x = 5

 \Rightarrow \frac{10}{6} \log_2 x = 5

 \Rightarrow \log_2 x = 3  \Rightarrow \log_2 x = \log_2 8 \Rightarrow x = 8

(v) \log (3x+2) + \log (3x-2) = 5 \log 2

 \Rightarrow \log (3x+2) (3x-2) = \log 2^5

 \Rightarrow 9x^2 - 4 = 32

 \Rightarrow 9x^2= 36

 \Rightarrow x^2 = 4 

 \Rightarrow x = 2

(vi) \log_3 (x+1) - 1 = 3 + \log_3 (x-1)

\Rightarrow \log_3 (x+1) - \log_3 3 = \log_3 3^3 + \log_3 (x-1)

\Rightarrow \log_3 (x+1) - \log_3 (x-1) = \log_3 3 + \log_3 3^3

\Rightarrow \log_3 \Big( \frac{x+1}{x-1} \Big) = \log_3 (3 \times 3^3)

\Rightarrow \frac{x+1}{x-1} = 81

\Rightarrow x+1 = 81x - 81

\Rightarrow or \ x = \frac{82}{80} = \frac{41}{40} 

(vii) \log_x 25 - \log_x 5 + \log_x \Big( \frac{1}{125} \Big) =2

\Rightarrow 2 \log_x 5 - \log_x 5 + \log_x (5^{-3}) = 2

\Rightarrow \log_x 5 - 3 \log_x 5 = \log_x x^2

\Rightarrow -2 \log_x 5 = \log_x x^2

\Rightarrow x^2 = 5^{-2}

\Rightarrow x^2 = ( \frac{1}{5} )^2

(viii) \log (x+1) + \log (x-1) = 3 \log 2 + \log 3, x >0

\Rightarrow \log (x^2 -1) = \log(2^3 \times 3)

\Rightarrow x^2 -1 = 24

\Rightarrow x^2 = 25

\Rightarrow x = 5

(ix) \log_3 x + \log_9 x + \log_{81} x = \frac{7}{4} 

\Rightarrow \log_3  x + \log_{3^2} x^1 + \log_{3^4} x^1 = \frac{7}{4} 

\Rightarrow \log_3 x + \frac{1}{2}  \log_3 x + \frac{1}{4}  \log_3 x = \frac{7}{4} 

\Rightarrow \log_3 x ( \frac{4+2+1}{4} ) = \frac{7}{4} 

\Rightarrow \log_3 x = 1 = \log_3 3

\Rightarrow x = 3

(x) \log_2 x + \log_8 x + \log_{32} x = \frac{23}{15} 

\Rightarrow \log_2 x + \log_{2^3} x^1 + \log_{2^5} x^1 = \frac{23}{15} 

\Rightarrow \log_2 x + \frac{1}{3} \log_2 x + \frac{1}{5} \log_2 x = \frac{23}{15} 

\Rightarrow \log_2 x \Big( 1 + \frac{1}{3} + \frac{1}{5} \Big) = \frac{23}{15} 

\Rightarrow \log_2 x ( \frac{23}{15} ) = \frac{23}{15} 

\Rightarrow \log_2 x = 1 = \log_2 2

\Rightarrow x = 2

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Question 22: If \frac{\log x}{\log 5} = \frac{\log y^2}{\log 2} = \frac{\log 9}{\log \frac{1}{3}} , find x and y

Answer:

Given \frac{\log x}{\log 5} = \frac{\log y^2}{\log 2} = \frac{\log 9}{\log \frac{1}{3}}

\Rightarrow \frac{\log x}{\log 5} = \frac{\log 9}{\log \frac{1}{3}} = \frac{\log 3^2}{\log 3^{-1}} = -2

\Rightarrow \log x = -2 \log 5 = \log 5^{-2} \Rightarrow x = \frac{1}{25}

\frac{\log y^2}{\log 2} = \frac{\log 9}{\log \frac{1}{3}}

\Rightarrow \frac{\log y^2}{\log 2} = \log 2 \frac{ \log 9}{\log 3^{-1}} 

\Rightarrow \frac{\log y^2}{\log 2} = \log 2 \frac{2 \log 3}{-1 . \log 3} = -2 \log 2 = \log \frac{1}{4} 

\Rightarrow \Rightarrow y^2 = \frac{1}{4} \Rightarrow y = \frac{1}{2} 

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Question 23: If a = \log_{10} 20 and b = \log_{10} 25 , find the value of x , if 2 \log_{10}(x+1)=2a-b

Answer:

Given 2a-b = 2 \log_{10}20 - \log_{10} 25 = \log_{10} \Big( \frac{20 \times 20}{25} \Big) = \log_{10} 4^2

2 \log_{10} (x+1) = (2a-b)

\log_{10} (x+1)^2 = \log_{10} 4^2

Therefore (x+1)^2 = 4^2 \Rightarrow x + 1 = 4 \Rightarrow x = 3

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Question 24: Given \log 2 = 0.3010 and \log 3 = 0.4771 , find the value of each of the following:

(i) \log 12 = \log (3 \times 4) = \log 3 + 2 \log 2 = 0.4771 + 2 \times 0.3010 = 1.0791

(ii) \log 5  = \log \frac{10}{2} = \log 10 - \log 2 = 1 - 0.3010 = 0.6990

(iii) \log 5^{\frac{1}{3}} = \frac{1}{3} (\log 10 - \log 2 ) = \frac{1}{3} (1 - 0.3010 ) = 0.2329

(iv) \log 108 = \log 27 + \log 4 = 3 \log 3 + 2 \log 2 = 3 \times 0.4771 + 2 \times 0.3010 = 2.034

(v) \log \Big( \frac{3}{8} \Big) = \log 3 - \log 8 = \log 3 - 2 \log 2 = 0.4771 - 3(0.3020) = -0.4259

(vi) \log 48 = \log 16 + \log 3 = 4 \log 2 + \log 3 = 4(0.3010) + 0.4771 = 1.6812

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Question 25: Without using \log tables show that

(i) \frac{\log \sqrt{27} + \log \sqrt{8} - \log \sqrt{125}}{\log 6 - \log 5} = \frac{3}{2} 

LHS = \frac{\log \sqrt{27} + \log \sqrt{8} - \log \sqrt{125}}{\log 6 - \log 5}

= \frac{\frac{3}{2} \log 3 + \frac{3}{2} \log 2 - \frac{3}{2} \log 5}{\log 6 - \log 5}

= \frac{3}{2} \frac{ \log \frac{3 \times 2}{ 5 } }{\log 6 - \log 5}

= \frac{3}{2} = RHS

(ii) \frac{\log \sqrt{27} + \log 8 - \log \sqrt{1000}}{\log 1.2 } = \frac{3}{2} 

= \frac{\frac{3}{2}\log 3 + \frac{3}{2}\log 2 - \frac{3}{2}\log 10 }{\log 1.2 }

= \frac{3}{2} (\frac{\log 3 + 2\log 2 - \log 10 }{\log 1.2 })

= \frac{3}{2} \log \frac{\frac{3 \times 4}{10} }{ \log 1.2}

= \frac{3}{2}

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