Logarithm and Antilogarithm Test 22

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Logarithm and Antilogarithm Test 22

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\frac { \log _{ 2 }{ 24 } }{ \log _{ 96 }{ 2 } } -\frac { \log _{ 2 }{ 192 } }{ \log _{ 12 }{ 2 } } $$ is:

A
3

B
0

C
2

D
1

Solution

Let $$\log _{ 2 }{ 12 }=a$$, then $$\frac { 1 }{ \log _{ 96 }{ 2 } }=\log _{ 2 }{ 96 }=\log _{ 2 }{ { 2 }^{ 3 } } \times 12=3+a$$
$$ \log _{ 2 }{ 24 }=1+a$$
$$\Rightarrow \log _{ 2 }{ 192 }=\log _{ 2 }{ \left( 16 \times 12 \right) }=4+a$$ and $$\frac { 1 }{ \log _{ 12 }{ 2 } } =\log _{ 2 }{ 12 } =a$$
Therefore, the given expression =$$(1+a)(3+a)-(4+a)a=3$$.
Question 2
1 / -0

$$a\times a\times b\times b\times b$$ can be written as

A
$$a^2b^3$$

B
$$a^3b^2$$

C
$$a^3b^3$$

D
$$a^5b^5$$

Solution

we have, $$a\times a\times b\times b\times b$$
$$=a^2 \times b^3 =a^2b^3$$

Hence, $$a\times a\times b\times b\times b$$ can be written as $$a^2b^3$$
Question 3
1 / -0

The value of $$0.2^{log_{\sqrt{5}} \Big( \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots \Big)}$$ is

A
$$4$$

B
$$log 4$$

C
$$log 2$$

D
none of these

Solution

$$\textbf{Given that:}$$

$$0.2^{\log_{\sqrt{5}} \left (\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + ... \right )} $$ $$....(1)$$

$$\textbf{Step 1: Let}\ y = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + ... $$

This is a infinite term G.P

where $$a = \dfrac{1}{4} , r = \dfrac{1}{2} , \left [S_{y} = \dfrac{a}{1 - r} \right ] $$

Hence sum of $$S_{y} = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}}$$

$$S_{y} = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} = \dfrac{1}{2} $$ $$....(2)$$

Put the this value in eq. $$(1)$$

$$ = (0.2)^{\log_{\sqrt{5}} \left (\dfrac{1}{2} \right )} $$

$$ = \left (\dfrac{1}{5} \right )^{\log_{5^{1/2}} \left (\dfrac{1}{2} \right )}$$ $$....(3)$$

$$\textbf{Step 2: Using the principle properties of logarithm} $$

$$\Rightarrow \log_{a^{P}} {m} = \dfrac{1}{p} \log_{a} m $$ [$$ a>0,\neq1 , m>0$$]

$$\Rightarrow a^{\log_{a} m} = m $$

So from eq. $$(3)$$

$$=\left (\dfrac{1}{5} \right )^{2 \log_{5} \left (\frac{1}{2} \right )} $$

$$=(5^{-1})^{2 \log_{5} \frac{1}{2}}$$

$$=5^{-2 \log_{5} \frac{1}{2}}$$

$$ =5^{\log_{5} \left (\frac{1}{2} \right )^{-2}} $$

$$ = \left (\dfrac{1}{2} \right )^{-2} $$

$$ = \dfrac{1}{2^{-2}} $$

$$ = 2^{2}$$

$$ = 4 $$
Question 4
1 / -0

$$\left(\dfrac {1}{10}\right)^0$$ is equal to

A
$$0$$

B
$$\dfrac {1}{10}$$

C
$$1$$

D
$$10$$

Solution

Using law of exponents, $$a^0=1$$
where $$a\neq 1$$
$$\Rightarrow \ \left(\dfrac {1}{10}\right)^0 =1$$
Question 5
1 / -0

By solving $$(6^0 -7^0) \times (6^0+7^0)$$, we get ________.

A
$$1$$

B
$$0$$

C
$$2$$

D
$$-1$$

Solution

$$0$$
$$(6^0 -7^0) \times (6^0+7^0)$$
$$=(1-1) \times (1+1)$$
$$=0\times 2=0$$
Question 6
1 / -0

$$\log_{\sqrt{2}} x = 4$$ then value of $$x$$ will be

A
$$4\sqrt{2}$$

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

C
$$4$$

D
$$4*\sqrt{2}$$

Solution

$$\because \log_{a}n=x$$
$$\therefore a^{x}=n$$
Given $$\log_{\sqrt{2}} x= 4$$
Then $$(\sqrt{2})^{4} = x$$
$$\Rightarrow x = (\sqrt{2})^{4} = (2^{1/2})^{4} = 2^{2} = 4$$
Hence, option (C) is correct.
Question 7
1 / -0

$$\log_{x} 243 = 2.5$$, then value of $$x$$ will be:

A
$$9$$

B
$$3$$

C
$$1$$

D
$$81$$

Solution

$$\because \log_{a} n = x$$
$$\Rightarrow a^{x} = n$$
Similarly $$\log_{x} 243 = 2.5$$
$$243 = x^{2.5}$$
$$x^{5/2} = 243 = (3)^{5}$$
Squaring on both sides,
$$(x^{5/2})^{2} = [(3)^{5}]^{2}$$
$$x^{5} = (3^{2})^{5}$$
On comparing, $$x = 3^{2} = 9$$
Hence, option (A) is correct
Question 8
1 / -0

The value of $$\log (1 + 2 * 3)$$:

A
$$2\log3$$

B
$$\log1.\log.2+\log3$$

C
$$\log1+\log2+\log3$$

D
$$\log7$$

Solution

$$\log (1 + 2 * 3) = \log (1 + 6) = \log 7$$
Hence, option (D) is correct
Question 9
1 / -0

Number $$\log_{2} 7$$ is:

A
Integer

B
Rational

C
Irrational

D
Prime

Solution

$$\log_{2} 7 = x$$
$$7 = 2^{x}$$
Taking $$\log$$ on both sides,
$$\log 7 = \log 2^{x}$$
$$\Rightarrow x \log 2 = \log 7$$
$$\Rightarrow x=\dfrac{\log 7}{\log 2}=\dfrac{0.8451}{0.3010}=2.8076$$
= Irrational number.
Hence, option (C) is correct.
Question 10
1 / -0

Multiply $$10^4$$ by $$10^2$$

A
$$10^8$$

B
$$10^2$$

C
$$10^6$$

D
$$10^{-2}$$

E
$$10^3$$

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Logarithm and Antilogarithm Test 22

Result

Logarithm and Antilogarithm Test 22

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SHARING IS CARING

Question 1

3

0

2

1

Solution

Question 2

$$a^2b^3$$

$$a^3b^2$$

$$a^3b^3$$

$$a^5b^5$$

Solution

Question 3

$$4$$

$$log 4$$

$$log 2$$

none of these

Solution

Question 4

$$0$$

$$\dfrac {1}{10}$$

$$1$$

$$10$$

Solution

Question 5

$$1$$

$$0$$

$$2$$

$$-1$$

Solution

Question 6

$$4\sqrt{2}$$

$$\frac{1}{4}$$

$$4$$

$$4*\sqrt{2}$$

Solution

Question 7

$$9$$

$$3$$

$$1$$

$$81$$

Solution

Question 8

$$2\log3$$

$$\log1.\log.2+\log3$$

$$\log1+\log2+\log3$$

$$\log7$$

Solution

Question 9

Integer

Rational

Irrational

Prime

Solution

Question 10

$$10^8$$

$$10^2$$

$$10^6$$

$$10^{-2}$$

$$10^3$$

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