Logarithm and Antilogarithm Test 15

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Logarithm and Antilogarithm Test 15

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Weekly Quiz Competition

Question 1
1 / -0

The logarithm form of $$\displaystyle (81)^\frac{3}{4} = 27$$ is $$\log_{81} 27 = \displaystyle \frac{3}{m}$$. Then value of $$m$$ is equal to

A
$$2$$

B
$$0$$

C
$$4$$

D
$$3$$

Solution

Given, $$\log_{81} 27 = \displaystyle \frac{3}{m}$$ ...(1)

Now, $$\displaystyle (81)^\frac{3}{4} = 27$$

Converting in logarithm form, we can write

$$\log_{81} 27 = \displaystyle \frac{3}{4}$$. ...(2)

From (1) and (2), we get

$$\displaystyle \dfrac{3}{m}=\displaystyle \dfrac{3}{4}$$

$$\therefore m=4$$
Question 2
1 / -0

The logarithm form of $$10^{-3} = 0.001$$ is $$\log_{10} 0.001 = -m$$, then value of $$m$$ is

A
$$-1$$

B
$$-2$$

C
$$3$$

D
$$-4$$

Solution

Given, $$ { 10 }^{ -3 }= 0.001$$
$$ \implies \log _{ 10 }{ 0.001 } = -3$$
$$\therefore m = 3$$.
Question 3
1 / -0

If $$\log_{10}(x - 10) = 1$$, then value of $$x$$ is

A
$$10$$

B
$$13$$

C
$$20$$

D
$$26$$

Solution

Since, $$ \log _{ 10 }{ (x-10) } =1$$

$$\implies (x -10)=10^1$$

$$ \implies x = 20 $$.
Question 4
1 / -0

If $$\displaystyle 64^{a}=\frac{1}{256^{b}}$$, then $$3a + 4b$$ equals:

A
$$2$$

B
$$4$$

C
$$8$$

D
$$0$$

Solution

$$ 64^{a}=\dfrac{1}{256^{b}}$$
$$\Rightarrow (2^{6})^{a}=\dfrac{1}{(2^{8})^{b}}$$
$$ \Rightarrow 2^{6a}\times 2^{8b}=1$$
$$\Rightarrow 2^{6a+8b}=2^{0}$$
$$ \Rightarrow 6a+8b=0$$
$$\Rightarrow 3a+4b=0$$
Question 5
1 / -0

If $$x = 2$$ and $$y = 3$$, then find the value of $$\left[ \displaystyle\frac { 1 }{ x^{ x } } +\displaystyle\frac { 1 }{ y^{ y } } \right] $$.

A
$$ \displaystyle\frac { -31 }{108 } $$

B
$$ \displaystyle\frac { 31 }{108 } $$

C
$$ \displaystyle\frac { 125 }{171} $$

D
$$ \displaystyle\frac { 153 }{222} $$

Solution

$$\cfrac { 1 }{ x^{ x } } +\cfrac { 1 }{ y^{ y } } =\cfrac { 1 }{ 2^{ 2 } } +\cfrac { 1 }{ 3^{ 3 } }$$
$$ =\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 27 } \\ =\cfrac { 27+4 }{ 108 } $$
$$=\cfrac { 31 }{ 108 } $$
Question 6
1 / -0

If $$\log_4m\,=\,1.5$$, then the value of $$m$$ is equal to

A
$$m=8$$

B
$$m=4$$

C
$$m=16$$

D
$$m=64$$

Solution

Since, $$\log _{ 4 } m\, =\, 1.5$$
$$\Rightarrow m={ 4 }^{ 1.5 }=2^{2\times 1.5}=2^3=8$$
Question 7
1 / -0

If $$ \log (a)^{3} + \log a$$ = $$m \log a$$, then value of $$m$$ is equal to

A
$$1$$

B
$$4$$

C
$$2$$

D
$$-5$$

Solution

$$\textbf{Step 1 : Use the rule of logarithm}$$
$$\text{According to logarithm power rule,}$$
$$\because \log_e( X ^Y) = Y . \log_e(X)$$
$$\text{Therefore, }\log(a)^3 = 3. \log \:a$$
$$\text{Since,} \log (a)^3 + \log a = m \log a$$
$$\text{Therefore, }3. \log a + \log a=m \log a$$
$$4 \log a=m \:\log\:a$$
$$\text{By comparing both the sides we get,}$$
$$m=4 $$

$$\textbf{Hence, option B is correct.}$$
Question 8
1 / -0

The value of $$4(7^1.\, 7^{-1}.\, 7^{-1}.7^0)$$ is

A
$$3\displaystyle \frac {7}{3}$$

B
$$\displaystyle \frac {6}{7}$$

C
$$3\displaystyle \frac {3}{7}$$

D
None of these

Solution
Question 9
1 / -0

If $$ \dfrac{\log 81}{\log 27} = x$$, then the value of $$x$$ is equal to

A
$$\dfrac { 4 }{ 3 } \\ \\ $$

B
$$\dfrac { 3 }{ 4 } \\ \\ $$

C
$$\dfrac { 1 }{ 3 } \\ \\ $$

D
Not solvable

Solution

$$ \dfrac{\log 81}{\log 27} = x$$
$$\Rightarrow x = \dfrac{\log 3^4}{\log 3^3} = \dfrac{4 \log 3}{3 \log 3} = \dfrac{4}{3}$$
Question 10
1 / -0

If $$\displaystyle \sqrt{3^{n}}=81$$. Then n is equal to

A
2

B
4

C
6

D
8

Solution

$$\Rightarrow \sqrt{3^{n}}=81$$
$$\Rightarrow 3^{n/2}=3^{4}$$
Base are same so
$$\Rightarrow \frac{n}{2}=4\Rightarrow n=8$$

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Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 15

Result

Logarithm and Antilogarithm Test 15

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SHARING IS CARING

Question 1

$$2$$

$$0$$

$$4$$

$$3$$

Solution

Question 2

$$-1$$

$$-2$$

$$3$$

$$-4$$

Solution

Question 3

$$10$$

$$13$$

$$20$$

$$26$$

Solution

Question 4

$$2$$

$$4$$

$$8$$

$$0$$

Solution

Question 5

$$ \displaystyle\frac { -31 }{108 } $$

$$ \displaystyle\frac { 31 }{108 } $$

$$ \displaystyle\frac { 125 }{171} $$

$$ \displaystyle\frac { 153 }{222} $$

Solution

Question 6

$$m=8$$

$$m=4$$

$$m=16$$

$$m=64$$

Solution

Question 7

$$1$$

$$4$$

$$2$$

$$-5$$

Solution

Question 8

$$3\displaystyle \frac {7}{3}$$

$$\displaystyle \frac {6}{7}$$

$$3\displaystyle \frac {3}{7}$$

None of these

Solution

Question 9

$$\dfrac { 4 }{ 3 } \\ \\ $$

$$\dfrac { 3 }{ 4 } \\ \\ $$

$$\dfrac { 1 }{ 3 } \\ \\ $$

Not solvable

Solution

Question 10

2

4

6

8

Solution

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