Logarithm and Antilogarithm Test 12

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Logarithm and Antilogarithm Test 12

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Question 1
1 / -0

The value of the expression $$\displaystyle\dfrac{(\log x - \log y)(\log x^{2} + \log y^{2})}{(\log x^{2} - \log y^{2})(\log x + \log y)}$$ is equal to

A
$$0$$

B
$$1$$

C
$$\log\displaystyle \frac{x}{y}$$

D
$$\log xy$$

Solution

Let $$E = \dfrac { (\log x-\log y)(\log { x }^{ 2 }+\log { y }^{ 2 }) }{ (\log { x }^{ 2 }-\log { y }^{ 2 })(\log x+\log y) } $$
Using the formula,
$$\log a^x = x\log a$$
Therefore, $$E=\dfrac { (\log x-\log y)(2\log x+2\log y) }{ (2\log x-2\log y)(\log x+\log y) }$$
$$\Rightarrow E =\dfrac { (\log x-\log y)\times 2\times (\log x+\log y) }{ 2\times (\log x-\log y)\times (\log x+\log y) }$$
$$\Rightarrow E= 1 $$
Hence, option B is correct.
Question 2
1 / -0

If $$ \log_{10} x = a$$ and $$ \log_{10} y = b$$, then $$10^{a-1}$$ in terms of $$x$$ will be

A
$$\displaystyle \frac{x}{7}$$

B
$$\displaystyle \frac{x}{8}$$

C
$$\displaystyle \frac{x}{9}$$

D
$$\displaystyle \frac{x}{10}$$

Solution

Given, $$log _{10}x=a, \log _{10}y=b$$
$$\implies 10^a=x$$
$$\implies 10^{a-1}=\dfrac{x}{10}$$
Hence, option D is correct.
Question 3
1 / -0

If $$ 3^{\log_{4}{x}}=27$$, then $$x$$ is equal to

A
$$16$$

B
$$64$$

C
$$27$$

D
$$\log_2 16$$

Solution

$$ 3^{\log_{4}{x}}=27$$
$$\Rightarrow { 3 }^{ \log _{ 4 }{ x } }={ 3 }^{ 3 }$$
Equating powers of $$3$$
$$\log _{ 4 }{ x } =3$$
Taking exponential of the above equation, we get
$${ 4 }^{ 3 }=x$$
$$\therefore x=64$$
Question 4
1 / -0

The value of $$\log_418$$ is equal to

A
a rational number

B
an irrational number

C
a prime number

D
none of these

Solution

$$log_{4}18$$

$$=\dfrac{log18}{log4}$$

$$=\dfrac{log2+log9}{2log2}$$

$$=\dfrac{log2+2log3}{2log2}$$

$$=\dfrac{log2}{2log2}+\dfrac{2log3}{2log2}$$

$$=\dfrac{1}{2}+log_{2}3$$
Since $$log_{2}3$$ is irrational.
Hence
$$\dfrac{1}{2}+log_{2}3$$ is irrational.
Therefore
$$log_{4}18$$ is an irrational number.
Question 5
1 / -0

Evaluate the expression by using logarithm tables: $$ \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$$

A
$$11.01$$

B
$$12.01$$

C
$$13.01$$

D
$$14.01$$

Solution

Let $$x= \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$$
Taking logarithm on both sides,
$$ \log { x } =\log { (17.42)^{ 2/{ 3 } } } +\log { 18.42 } -\log { \sqrt { 126.37 } } $$

$$\log { x } =\dfrac { 2 }{ 3 } \log { 17.42 } +\log { 18.42 } -\dfrac { 1 }{ 2 } \log { (126.37) } $$

$$ \log { x } =\dfrac { 2 }{ 3 } \log { (1.742\times 10) } +\log { (1.842\times 10) } -\dfrac { 1 }{ 2 } \log { (1.264\times { 10 }^{ 2 }) } $$

$$ \log { x } =\dfrac { 2 }{ 3 } { (1.2410) } + { 1.2653 } -\dfrac { 1 }{ 2 } { (2.1018) } $$

$$\log { x } =0.8273+1.2653-1.0509$$

$$\log { x } =1.0417$$
$$\Rightarrow x= \text{antilog }(1.0417)$$
$$\Rightarrow x = 11.01$$
Question 6
1 / -0

If $$ \log_{10} x = a$$ and $$ \log_{10} y = b$$, then $$10^{2b}$$ in terms of $$y$$ is equal to

A
$$y$$

B
$$y^2$$

C
$$y^3$$

D
$$y^4$$

Solution

Given, $$\log _{10}x=a, \log _{10}y=b$$
$$ \Rightarrow 10^b=y$$
Consider, $$ 10^{2b}$$
$$=10^{b+b}$$
$$=10^b.10^b$$
$$=y.y$$
$$=y^2$$.
Question 7
1 / -0

The number $$\displaystyle N= 6 \log_{10}2+\log _{10}31 $$ lies between two successive integers whose sum is equal to:

A
5

B
7

C
9

D
10

Solution

$$N=\log _{ 10 }{ 64 } +\log _{ 10 }{ 31 } =\log _{ 10 }{ 1984 } $$
Now, we know that $$\log_{10}{1000}=3$$ and $$\log_{10}{10000}=4$$.
So, $$1000<1984<10000$$
$$\Rightarrow\log_{10}{1000}<\log_{10}{1984}<\log_{10}{10000}$$
$$\Rightarrow3<\log_{10}{1984}<4$$
$$\Rightarrow3<\log _{ 10 }{ 64 } +\log _{ 10 }{ 31 }<4$$
So, it lies between two successive integers $$3$$ and $$4$$.
So, their sum is equal to $$3+4=7$$
Hence. option $$B$$ is correct.
Question 8
1 / -0

If $$\log 3 = 0.477$$ and $$(1000)^x = 3$$, then the value of $$x$$ will be

A
$$0.159$$

B
$$0.62$$

C
$$0.162$$

D
$$0.59$$

Solution

Since, $$(1000)^x=3$$
$$\therefore \log [(1000)^x] = \log3$$
$$\Rightarrow x \log 1000 = \log 3$$
$$\Rightarrow x \log (10^3) = \log 3$$
$$\Rightarrow x \cdot 3\log 10 = \log 3$$
$$\Rightarrow 3x = \dfrac {\log 3}{\log 10}$$
$$\Rightarrow x = \dfrac{0.477}{3 \times 1} = 0.159$$
Question 9
1 / -0

If $$\log_{5} x = y$$, then value of $$5^{5y}$$ is equal to

A
$$\dfrac{x}{5}$$

B
$$5x$$

C
$$\log_{x}5$$

D
$$x^{5}$$

Solution

$$\log_{5} x = y \Rightarrow 5^{y} = x$$
$$\therefore (5^{y})^5 = x^5 \Rightarrow 5^{5y} = x^{5}$$
Question 10
1 / -0

The value of $$x$$ satisfying the logarithm $$\log_{243} x = 0.8$$ is equal to

A
$$81$$

B
$$1.8$$

C
$$2.43$$

D
$$27$$

Solution

Given, $$\log_{243} x = 0.8$$
$$ \Rightarrow 243^{0.8} = x$$
$$\Rightarrow (3^5)^{4/5} = x$$
$$ \Rightarrow 3^4 = x$$
$$\Rightarrow x = 81$$

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Logarithm and Antilogarithm Test 12

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Logarithm and Antilogarithm Test 12

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$\log\displaystyle \frac{x}{y}$$

$$\log xy$$

Solution

Question 2

$$\displaystyle \frac{x}{7}$$

$$\displaystyle \frac{x}{8}$$

$$\displaystyle \frac{x}{9}$$

$$\displaystyle \frac{x}{10}$$

Solution

Question 3

$$16$$

$$64$$

$$27$$

$$\log_2 16$$

Solution

Question 4

a rational number

an irrational number

a prime number

none of these

Solution

Question 5

$$11.01$$

$$12.01$$

$$13.01$$

$$14.01$$

Solution

Question 6

$$y$$

$$y^2$$

$$y^3$$

$$y^4$$

Solution

Question 7

5

7

9

10

Solution

Question 8

$$0.159$$

$$0.62$$

$$0.162$$

$$0.59$$

Solution

Question 9

$$\dfrac{x}{5}$$

$$5x$$

$$\log_{x}5$$

$$x^{5}$$

Solution

Question 10

$$81$$

$$1.8$$

$$2.43$$

$$27$$

Solution

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