Logarithm and Antilogarithm Test 16

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Logarithm and Antilogarithm Test 16

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$(-3)^0 - (-3)^3 - (-3)^{-1} + (-3)^4 - (-3)^{-2}$$ is

A
$$109\, \displaystyle \frac {2}{9}$$

B
$$109\, \displaystyle \frac {9}{2}$$

C
$$109$$

D
None of these

Solution

Given expression is $$(-3)^0 - (-3)^3 - (-3)^{-1} + (-3)^4 - (-3)^{-2}$$

Using the law of exponent, $$a^{-m}=\dfrac{1}{a^m}$$, we can write it as:

$$=1 - (-27) - \left(-\cfrac {1}{3}\right)$$ $$+ 81 - \cfrac 19$$

$$= 1 + 27 + \cfrac {1}{3} + 81 - \cfrac {1}{9}$$

$$=109 + \cfrac {2}{9}\, $$

$$=\, 109 \displaystyle \frac {2}{9}$$

Hence, option A is correct.
Question 2
1 / -0

The value of $$512^\frac {-2}{9}$$ is

A
$$\displaystyle \frac {1}{2}$$

B
$$2$$

C
$$4$$

D
$$\displaystyle \frac {1}{4}$$

Solution

$$\begin{aligned}{}{\left( {512} \right)^{ - {\textstyle{2 \over 9}}}} &= {\left( {{2^9}} \right)^{ - {\textstyle{2 \over 9}}}}\\&={2^{9 \times \frac{{ - 2}}{9}}} \quad\quad\quad\because[(a^b)^c=a^{bc}]\\&= {2^{ - 2}}\\ &= \frac{1}{4}\end{aligned}$$

Hence, option $$D$$ is correct.
Question 3
1 / -0

$$\log_4 $$1 is equal to

A
$$1$$

B
$$0$$

C
$$\infty$$

D
none of these

Solution

$$log_a 1 = m$$
$$\Rightarrow a^m = 1 a^0 \Rightarrow m = 0$$
Question 4
1 / -0

Value of $$\displaystyle \log _{4}18 $$ is:

A
an irrational number

B
a rational number

C
natural number

D
whole number

Solution

$$\log_{4}{18}=\log_{2^{2}}{(2.3^{2})}=\dfrac{1}{2}\log_{2}(2.3^{2})$$
$$=\dfrac{1}{2}\left [ \log_{2}{2}+\log_{2}{3^{2}} \right ]=\dfrac{1}{2}\left [ 1+2.\log_{2}{3} \right ]$$

$$\log_{2}{3}$$ is an irrational number.

Hence, $$\dfrac{1}{2}\left [ 1+2.\log_{2}{3} \right ]$$ is also an irrational number.
Question 5
1 / -0

The value of $$[ 5 (8^{\tfrac 13} + 27^{\tfrac 13} )^3 ]^{\tfrac 14}$$ is

A
$$5^4$$

B
$$5^{\tfrac 14}$$

C
$$5$$

D
None of these

Solution

$$\left [ 5 \left ( 2^{3\times \tfrac {1}{3}}+ 3^{3\times \tfrac {1}{3}} \right )^3 \right ]^{\tfrac 14}$$

$$=[ 5 (2 + 3)^3 ] ^{\tfrac 14}$$

$$= (5 \times 5^3)^{\tfrac 14} = 5^{4\times \tfrac {1}{4}} = 5$$
Question 6
1 / -0

Value of the expression $$\displaystyle \log _{2}\sqrt[5]{2.\sqrt[3]{2\sqrt{2}}}$$ is

A
$$0.1$$

B
$$0.2$$

C
$$0.3$$

D
does not exist

Solution

$$\log_{2}\left ( \sqrt[5]{2.\sqrt[3]{2\sqrt{2}}} \right )=$$
$$=\log_{2}\left ( \sqrt[5]{2.({2^{3/2}})^{1/3}} \right )$$
$$=\log_{2}\left ( \sqrt[5]{2.2^{1/2}} \right )$$
$$=\log_{2}\left ( 2^{3/2} \right )^{1/5}=\dfrac{1}{5}\, \log_{2}.{2^{3/2}}$$
$$=\dfrac{3}{10}.\log_{2}{2}=\dfrac{3}{10}=0.3$$
Question 7
1 / -0

The value of $$(6^{-1} -7^{-1})^{-1} (5^{-1} -4^{-1})^{-1}$$ is equal to

A
$$\displaystyle \frac{1}{2}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{1}{4}$$

D
None of these

Solution

Given expression is $$\left({ 6 }^{ -1 }-7^{ -1 }\right)^{ -1 }\left(5^{ -1 }-4^{ -1 }\right)^{ -1 }$$.

From laws of exponents, we know that $$a^{-m}=\left(\dfrac{1}{a}\right)^m$$

Hence, the given expression can be simplified as:
$$\left(\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 } \right)^{ -1 }\left(\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 4 } \right)^{ -1 }$$

Solving the fractions, we get:
$$\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 }=\cfrac{1}{42}$$
and
$$\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 4 }=\cfrac{-1}{20}$$

Hence, the expression simplifies to:
$$\left(\cfrac { 1 }{ 42 } \right)^{ -1 }\left(-\cfrac { 1 }{ 20 } \right)^{ -1 }$$

Again applying the law of exponent, $$a^{-m}=\left(\dfrac{1}{a}\right)^m$$, we get:

$$(42)^1(-20)^1$$

$$\Rightarrow 42\times (-20)=-840$$

Hence, $$\left({ 6 }^{ -1 }-7^{ -1 }\right)^{ -1 }\left(5^{ -1 }-4^{ -1 }\right)^{ -1 }=-840$$.
Question 8
1 / -0

If $$\left (\dfrac {a}{b}\right )^{x-1}=\left (\dfrac {a}{b}\right )^{x-3}$$ then the value of $$x$$ is

A
-1

B
1

C
2

D
3

Solution

Given,

$$\left (\dfrac {a}{b}\right )^{x-1}=\left (\dfrac {a}{b}\right )^{-(x-3)}$$

as the terms are same we can equate the powers/exponents of the terms

$$\Rightarrow x-1=-(x-3)$$

$$\Rightarrow x-1=-x+3$$

$$\Rightarrow 2x=4$$

$$\Rightarrow x=2$$
Question 9
1 / -0

SImplify: $$(256)^{0.16} \times (256)^{0.09}$$

A
$$4$$

B
$$16$$

C
$$64$$

D
$$256.25$$

Solution

$$(256)^{0.16} \times (256)^{0.09}$$

$$=(256)^{(0.16 + 0.09)}$$

$$= (256)^{0.25} = (256)^{\tfrac {25}{100}} = (256)^{\tfrac 14}$$

$$= (4^4)^{\tfrac 14} = 4^1 = 4$$
Question 10
1 / -0

The value of $$[10^{150}\, \div\, 10^{146} ]$$ is

A
$$1000$$

B
$$10000$$

C
$$100000$$

D
$$10^5$$

Solution

$$(10)^{150} \div (10)^{146} = \cfrac {(10)^{150}}{(10)^{146}}$$ $$= 10^{(150 - 146)} = 10^4 = 10000$$

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Logarithm and Antilogarithm Test 16

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Logarithm and Antilogarithm Test 16

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Question 1

$$109\, \displaystyle \frac {2}{9}$$

$$109\, \displaystyle \frac {9}{2}$$

$$109$$

None of these

Solution

Question 2

$$\displaystyle \frac {1}{2}$$

$$2$$

$$4$$

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

Solution

Question 3

$$1$$

$$0$$

$$\infty$$

none of these

Solution

Question 4

an irrational number

a rational number

natural number

whole number

Solution

Question 5

$$5^4$$

$$5^{\tfrac 14}$$

$$5$$

None of these

Solution

Question 6

$$0.1$$

$$0.2$$

$$0.3$$

does not exist

Solution

Question 7

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{1}{3}$$

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

None of these

Solution

Question 8

-1

1

2

3

Solution

Question 9

$$4$$

$$16$$

$$64$$

$$256.25$$

Solution

Question 10

$$1000$$

$$10000$$

$$100000$$

$$10^5$$

Solution

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