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21. If $\log_{10000} x = - \frac{1}{4},$ then the value of x is = ?

A. $\frac{1}{10}$

B. $\frac{1}{100}$

C. $\frac{1}{1000}$

D. $\frac{1}{10000}$

Discuss

Solution:

\begin{aligned} let \log_{10000} x = - \frac{1}{4} \\ \Rightarrow x = {(10000)}^{- \frac{1}{4}} \\ = {(10^{4})}^{- \frac{1}{4}} \\ = 10^{- 1} \\ = \frac{1}{10} \end{aligned}

22. $\frac{\log \sqrt{8}}{\log 8} is equal to = ?$

A. $\frac{1}{\sqrt{8}}$

B. $\frac{1}{4}$

C. $\frac{1}{2}$

D. $\frac{1}{8}$

Discuss

Solution:

\begin{aligned} \frac{\log \sqrt{8}}{\log 8} \\ = \frac{\log {(8)}^{\frac{1}{2}}}{\log 8} \\ = \frac{\frac{1}{2} \log 8}{\log 8} \\ = \frac{1}{2} \end{aligned}

23. The value of $\frac{6 \log_{10} 1000}{3 \log_{10} 100}$ is equal to -

A. 0

B. 1

C. 2

D. 3

Discuss

Solution:

\begin{aligned} \frac{6 \log_{10} 1000}{3 \log_{10} 100} \\ = \frac{6 \log_{10} 10^{3}}{3 \log_{10} 10^{2}} \\ = \frac{6 \times 3 \log_{10} 10}{3 \times 2 \log_{10} 10} \\ = \frac{18}{6} \\ = 3 \end{aligned}

24. If $\log_{2} [\log_{3} (\log_{2} x)] = 1,$ then x is equal to = ?

A. 0

B. 12

C. 128

D. 512

Discuss

Solution:

\begin{aligned} \log_{2} [\log_{3} (\log_{2} x)] = 1 \\ \Rightarrow \log_{3} (\log_{2} x) = 2^{1} = 2 \\ \Rightarrow \log_{2} x = 3^{2} = 9 \\ \Rightarrow x = 2^{9} = 512 \end{aligned}

25. What is the value of the following expression?
$\log (\frac{9}{14}) -$ $\log (\frac{15}{16}) +$ $\log (\frac{35}{24})$

A. 0

B. 1

C. 2

D. 3

Discuss

Solution:

\begin{aligned} \log (\frac{9}{14}) - \log (\frac{15}{16}) + \log (\frac{35}{24}) \\ = \log (\frac{9}{14} \div \frac{15}{16} \times \frac{35}{24}) \\ = \log (\frac{9}{14} \times \frac{16}{15} \times \frac{35}{24}) \\ = \log 1 \\ = 0 \end{aligned}

26. $2 {\log_{10}}^{5} +$ $\log_{10} 8 -$ $\frac{1}{2} \log_{10} 4$ = ?

A. 2

B. 4

C. $2 - 2 lo {g_{10}}^{2}$

D. $4 - 4 lo {g_{10}}^{2}$

Discuss

Solution:

\begin{aligned} 2 \log_{10} 5 + \log_{10} 8 - \frac{1}{2} \log_{10} 4 \\ = \log_{10} (5^{2}) + \log_{10} 8 - \log_{10} (4^{\frac{1}{2}}) \\ = \log_{10} 25 + \log_{10} 8 - \log_{10} 2 \\ = \log_{10} (\frac{25 \times 8}{2}) \\ = \log_{10} 100 \\ = 2 \end{aligned}

27. If $\log_{a} (a b) = x,$ then $\log_{b} (a b)$ is -

A. $\frac{1}{x}$

B. $\frac{x}{x + 1}$

C. $\frac{x}{1 - x}$

D. $\frac{x}{x - 1}$

Discuss

Solution:

\begin{aligned} lo g_{a} (a b) = x \\ \Rightarrow \frac{\log a b}{\log a} = x \\ \Rightarrow \frac{\log a + \log b}{\log a} = x \\ \Rightarrow 1 + \frac{\log b}{\log a} = x \\ \Rightarrow \frac{\log b}{\log a} = x - 1 \\ \Rightarrow \frac{\log a}{\log b} = \frac{1}{x - 1} \\ \Rightarrow 1 + \frac{\log a}{\log b} = 1 + \frac{1}{x - 1} \\ \Rightarrow \frac{\log b}{\log b} + \frac{\log a}{\log b} = \frac{x}{x - 1} \\ \Rightarrow \frac{\log b + \log a}{\log b} = \frac{x}{x - 1} \\ \Rightarrow \frac{log (a b)}{\log b} = \frac{x}{x - 1} \\ \Rightarrow lo g_{b} (a b) = \frac{x}{x - 1} \end{aligned}

28. If $\log_{10} 2 = a$ and $\log_{10} 3 = b,$ then $\log_{5} 12$ = ?

A. $\frac{a + b}{1 + a}$

B. $\frac{2 a + b}{1 + a}$

C. $\frac{a + 2 b}{1 + a}$

D. $\frac{2 a + b}{1 - a}$

Discuss

Solution:

\begin{aligned} \log_{5} 12 = \log_{5} (3 \times 4) \\ = \log_{5} 3 + \log_{5} 4 \\ = \log_{5} 3 + 2 \log_{5} 2 \\ = \frac{\log_{10} 3}{\log_{10} 5} + \frac{2 \log_{10} 2}{\log_{10} 5} \end{aligned}

= \frac{\log_{10} 3}{\log_{10} 10 - \log_{10} 2}

+ \frac{2 \log_{10} 2}{\log_{10} 10 - \log_{10} 2}

\begin{aligned} = \frac{b}{1 - a} + \frac{2 a}{1 - a} \\ = \frac{2 a + b}{1 - a} \end{aligned}

29. If $\log_{10} 125 + \log_{10} 8 = x,$ then x is equal to -

A. $\frac{1}{3}$

B. 0.064

C. -3

D. 3

Discuss

Solution:

\begin{aligned} \log_{10} 125 + \log_{10} 8 = x \\ \Rightarrow \log_{10} (125 \times 8) = x \\ \Rightarrow x = \log_{10} (1000) \\ \Rightarrow x = \log_{10} {(10)}^{3} \\ \Rightarrow x = 3 \log_{10} 10 \\ \Rightarrow x = 3 \end{aligned}

30. If $\log_{5} (x^{2} + x) -$ $\log_{5} (x + 1)$ = 2, then the value of x is -

A. 5

B. 10

C. 25

D. 32

Discuss

Solution:

\begin{aligned} \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2 \\ \Rightarrow \log_{5} (\frac{x^{2} + x}{x + 1}) = 2 \\ \Rightarrow \log_{5} [\frac{x (x + 1)}{x + 1}] = 2 \\ \Rightarrow \log_{5} x = 2 \\ \Rightarrow x = 5^{2} \\ = 25 \end{aligned}

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