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31. If $\log \frac{a}{b} + \log \frac{b}{a} =$ $\log (a + b),$ then -

A. a + b = 1

B. a - b = 1

C. a = b

D. a² - b² = 1

Discuss

Solution:

\begin{aligned} \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b) \\ \Rightarrow \log (a + b) = \log (\frac{a}{b} \times \frac{b}{a}) \\ \Rightarrow \log (a + b) = \log 1 \\ S o, a + b = 1 \end{aligned}

32. $\log (\frac{a^{2}}{b c}) +$ $\log (\frac{b^{2}}{a c}) +$ $\log (\frac{c^{2}}{a b})$ is equal to -

A. 0

B. 1

C. 2

D. abc

Discuss

Solution:

\begin{aligned} Given Expression \\ = \log (\frac{a^{2}}{b c} \times \frac{b^{2}}{a c} \times \frac{c^{2}}{a b}) \\ = \log 1 \\ = 0 \end{aligned}

33. $\frac{1}{\log_{a} b} \times \frac{1}{\log_{b} c} \times \frac{1}{\log_{c} a}$ is equal to -

A. a + b + c

B. abc

C. 0

D. 1

Discuss

Solution:

\begin{aligned} Given Expression \\ (\frac{\log a}{\log b} \times \frac{\log b}{\log c} \times \frac{\log c}{\log a}) \\ = 1 \end{aligned}

34. $\frac{1}{(\log_{a} b c) + 1} +$ $\frac{1}{(\log_{b} c a) + 1} +$ $\frac{1}{(\log_{c} a b) + 1}$ is equal to -

A. 1

B. $\frac{3}{2}$

C. 2

D. 3

Discuss

Solution:

\begin{aligned} Given Expression \\ = \frac{1}{\log_{a} b c + \log_{a} a} + \frac{1}{\log_{b} c a + \log_{b} b} + \frac{1}{\log_{c} a b + \log_{c} c} \\ = \frac{1}{\log_{a} (a b c)} + \frac{1}{\log_{b} (a b c)} + \frac{1}{\log_{c} (a b c)} \\ = \log_{a b c} a + \log_{a b c} b + \log_{a b c} c \\ = \log_{a b c} (a b c) \\ = 1 \end{aligned}

35. If $\log_{10} 7 = a,$ then $\log_{10} (\frac{1}{70})$ is equal to -

A. -(1 + a)

B. (1 + a)^-1

C. $\frac{a}{10}$

D. $\frac{1}{10 a}$

Discuss

Solution:

\begin{aligned} \log_{10} (\frac{1}{70}) \\ = \log_{10} 1 - \log_{10} 70 \\ = - \log_{10} (7 \times 10) \\ = - (\log_{10} 7 + \log_{10} 10) \\ = - (a + 1) \end{aligned}

36. If $\log x - 5 \log 3 = - 2,$ then x equals -

A. 0.8

B. 0.81

C. 1.25

D. 2.43

Discuss

Solution:

\begin{aligned} \log x - 5 \log 3 = - 2 \\ \Rightarrow \log x - \log 3^{5} = - 2 \\ \Rightarrow \log (\frac{x}{3^{5}}) = - 2 \\ \Rightarrow \frac{x}{243} = 10^{- 2} = \frac{1}{100} \\ \Rightarrow x = \frac{243}{100} = 2.43 \end{aligned}

37. If $a = b^{2} = c^{3} = d^{4},$ then the value of $\log_{a} (a b c d)$ would be -

A. $\log_{a} 1 + \log_{a} 2 + \log_{a} 3 + \log_{a} 4$

B. $\log_{a} 24$

C. $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$

D. $1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}$

Discuss

Solution:

\begin{aligned} a = b^{2} = c^{3} = d^{4} \\ \Rightarrow b = a^{\frac{1}{2}}, c = a^{\frac{1}{3}}, d = a^{\frac{1}{4}} \\ ∴ \log_{a} (a b c d) \\ = \log_{a} (a \times a^{\frac{1}{2}} \times a^{\frac{1}{3}} \times a^{\frac{1}{4}}) \\ = \log_{a} a^{(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})} \\ = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) \log_{a} a \\ = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \end{aligned}

38. If $\log_{3} x + \log_{9} x^{2} + \log_{27} x^{3}$ $= 9,$ then x equals to -

A. 3

B. 9

C. 27

D. None of these

Discuss

Solution:

\begin{aligned} \Rightarrow \log_{3} x + \log_{9} x^{2} + \log_{27} x^{3} = 9 \\ \Rightarrow \log_{3} x + \log_{3^{2}} x^{2} + \log_{3^{3}} x^{3} = 9 \\ \Rightarrow \log_{3} x + \frac{2}{2} \log_{3} x + \frac{3}{3} \log_{3} x = 9 \\ \Rightarrow 3 \log_{3} x = 9 \\ \Rightarrow \log_{3} x = 3 \\ \Rightarrow x = 3^{3} = 27 \end{aligned}

39. If $\log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x})$ $= 0,$ what is the value of x ?

A. 2

B. 3

C. 4

D. 5

Discuss

Solution:

\begin{aligned} \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0 \\ \Rightarrow \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 7^{0} = 1 \\ \Rightarrow \sqrt{x + 5} + \sqrt{x} = 5^{1} = 5 \\ \Rightarrow {(\sqrt{x + 5} + \sqrt{x})}^{2} = 25 \\ \Rightarrow (x + 5) + x + 2 \sqrt{x + 5} \sqrt{x} = 25 \\ \Rightarrow 2 x + 2 \sqrt{x} \sqrt{x + 5} = 20 \\ \Rightarrow \sqrt{x} \sqrt{x + 5} = 10 - x \\ \Rightarrow x (x + 5) = {(10 - x)}^{2} \\ \Rightarrow x^{2} + 5 x = 100 + x^{2} - 20 x \\ \Rightarrow 25 x = 100 \\ \Rightarrow x = 4 \end{aligned}

40. If $\log 2 = 0.3010$ and $\log 3 = 0.4771,$ the value of $\log_{5} 512$ = ?

A. 2.870

B. 2.967

C. 3.876

D. 3.912

Discuss

Solution:

\begin{aligned} \log_{5} 512 \\ = \frac{\log 512}{\log 5} \\ = \frac{\log 2^{9}}{\log (\frac{10}{2})} \\ = \frac{9 \log 2}{\log 10 - \log 2} \\ = \frac{(9 \times 0.3010)}{1 - 0.3010} \\ = \frac{2.709}{0.699} \\ = \frac{2709}{699} \\ = 3.876 \end{aligned}

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