Practice | Hashtable

1. Which of the following statements is not correct?

A. log₁₀ 10 = 1

B. log (2 + 3) = log (2 x 3)

C. log₁₀ 1 = 0

D. log (1 + 2 + 3) = log 1 + log 2 + log 3

Solution:

\begin{aligned} A  =  Since lo g_{a} a = 1, so lo g_{10} 10 = 1 \\ B = log (2 + 3) = 5 \\ and log (2 \times 3) = log 6 \\ = log 2 + log 3 \\ ∴ log (2 + 3) \neq log (2 \times 3) . \\ C = Since lo g_{a} 1 = 0, s o lo g_{10} 1 = 0. \\ D  =  log (1 + 2 + 3) = log 6 \\ = log (1 \times 2 \times 3) \\ = log 1 + log 2 + log 3. \\ So (B) is incorrect \end{aligned}

2. If log 2 = 0.3010 and log 3 = 0.4771, the value of log₅ 512 is:

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}

3. $\frac{\log \sqrt{8}}{\log 8}$ is equal to:

A. $\frac{1}{6}$

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

4. If log 27 = 1.431, then the value of log 9 is:

A. 0.934

B. 0.945

C. 0.954

D. 0.958

Discuss

Solution:

\begin{aligned} \log 27 = 1.431 \\ \Rightarrow \log (3^{3}) = 1.431 \\ \Rightarrow 3 \log 3 = 1.431 \\ \Rightarrow \log 3 = 0.477 \\ ∴ \log 9 = \log (3^{2}) \\ = 2 \log 3 \\ = 2 \times 0.477 \\ = 0.954 \end{aligned}

5. $If log \frac{a}{b} + log \frac{b}{a} = log (a + b),$ 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}) = \log 1 \\ S o, a + b = 1 \end{aligned}

6. 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}

7. If log₁₀ 2 = 0.3010, then log₂ 10 is equal to:

A. $\frac{699}{301}$

B. $\frac{1000}{301}$

C. 0.3010

D. 0.6990

Discuss

Solution:

\begin{aligned} \log_{2} 10 \\ = \frac{1}{\log_{10} 2} \\ = \frac{1}{0.3010} \\ = \frac{10000}{3010} \\ = \frac{1000}{301} \end{aligned}

8. If log₁₀ 2 = 0.3010, the value of log₁₀ 80 is:

A. 1.6020

B. 1.9030

C. 3.9030

D. None of these

Discuss

Solution:

\begin{aligned} \log_{10} 80 \\ = \log_{10} (8 \times 10) \\ = \log_{10} 8 + \log_{10} 10 \\ = \log_{10} (2^{3}) + 1 \\ = 3 \log_{10} 2 + 1 \\ = (3 \times 0.3010) + 1 \\ = 1.9030 \end{aligned}

9. If $\log_{10} 5 + \log_{10} (5 x + 1)$ = $\log_{10}$ $(x + 5)$ $+$ $1,$ then x is equal to :

A. 1

B. 3

C. 5

D. 10

Discuss

Solution:

\begin{aligned} \Rightarrow \log_{10} 5 + \log_{10} (5 x + 1) = \log_{10} (x + 5) + 1 \\ \Rightarrow \log_{10} [5 (5 x + 1)] = \log_{10} [10 (x + 5)] \\ \Rightarrow 5 (5 x + 1) = 10 (x + 5) \\ \Rightarrow 5 x + 1 = 2 x + 10 \\ \Rightarrow 3 x = 9 \\ \Rightarrow x = 3 \end{aligned}

10. The value of $\frac{1}{\log_{3} 60} +$ $\frac{1}{\log_{4} 60} +$ $\frac{1}{\log_{5} 60}$ is :

A. 0

B. 1

C. 5

D. 60

Discuss

Solution:

Given expression

\begin{aligned} = \log_{60} 3 + \log_{60} 4 + \log_{60} 5 \\ = \log_{60} (3 \times 4 \times 5) \\ = \log_{60} 60 \\ = 1 \end{aligned}