Logarithm and Antilogarithm Test 4

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Logarithm and Antilogarithm Test 4

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$4$$

B
$$16$$

C
$$64$$

D
$$256.25$$

Solution

$${ \left( 256 \right) }^{ 0.16 }\times { \left( 256 \right) }^{ 0.09 }$$ $$={ \left( 256 \right) }^{ 0.16+0.09 }$$

$$={ \left( 256 \right) }^{ 0.25 }$$

$$={ \left( 256 \right) }^{ \tfrac { 25 }{ 100 } }$$

$$ ={ \left( 256 \right) }^{ \tfrac { 1 }{ 4 } }$$

$$={ 2 }^{ 8\times \tfrac { 1 }{ 4 } }$$ $$={ 4 }$$
Question 2
1 / -0

The value of $$\displaystyle \frac {1}{(216)^{-2/3}}\, +\, \displaystyle \frac {1}{(256)^{-3/4}}\, +\, \displaystyle \frac {1}{(32)^{-1/5}}$$ is:

A
$$102$$

B
$$105$$

C
$$107$$

D
$$109$$

Solution

$$ \cfrac {1}{(216)^{-\tfrac 23}} + \cfrac {1}{(256)^{-\tfrac 34}} + \cfrac {1}{(32)^{-\tfrac 15}}$$
$$= \cfrac {1}{(6^3)^{-\tfrac 23}} + \cfrac {1}{(4^4)^{-\tfrac 34}} + \cfrac {1}{(2^5)^{-\tfrac 15}}$$
$$= \cfrac {1}{6^{-2}} + \cfrac {1}{4^{-3}} + \cfrac {1}{2^{-1}}$$
$$ = (6^2 + 4^3 + 2^1)$$
$$= (36 + 64 + 2^1) $$
$$= 102$$
Question 3
1 / -0

The value of $$(8^{-25}\, -\, 8^{-26})$$ is

A
$$7\, \times\, 8^{-25}$$

B
$$7\, \times\, 8^{-26}$$

C
$$8\, \times\, 8^{-26}$$

D
None of these

Solution

$$(8^{-25} - 8^{-26})$$ $$=\left ( { \dfrac {1}{8^{25}}} - { \dfrac {1}{8^{26}}} \right ) $$

$$=\dfrac{8^{25}(8-1)}{8^{25}\times 8^{26}}$$

$$= \cfrac {(8-1)}{8^{26}}$$

$$=7 \times 8^{-26}$$

Thus option $$(B)$$ is correct.
Question 4
1 / -0

Evaluate: $$(0.04)^{-1.5} $$

A
$$25$$

B
$$125$$

C
$$250$$

D
$$625$$

Solution

$$(0.04)^{-1.5} = \left ( \cfrac {4}{100} \right )^{-1.5}$$$$=\,\left (\cfrac {1}{25} \right )^{-\tfrac 32} $$$$= (25)^{\tfrac 32}$$$$ = (5^2)^{\tfrac 32}$$$$= 5^3 = 125$$
Question 5
1 / -0

$$\displaystyle 2^{73}-2^{72}-2^{71}$$ is the same as :

A
$$\displaystyle 2^{69}$$

B
$$\displaystyle 2^{70}$$

C
$$\displaystyle 2^{71}$$

D
$$\displaystyle 2^{72}$$

Solution

$$2^{73}-2^{72}-2^{71}=2^{71}(2^2-2^1-1)$$

$$=2^{71}(4-2-1)$$

$$=2^{71}$$

Therefore, Option C is correct.
Question 6
1 / -0

if $$\dfrac{(23)^{9}-(23)^{8}}{22}=(23)^{x}$$ then $$x$$ equals :

A
$$7$$

B
$$8$$

C
$$9$$

D
$$1$$

Solution

$$LHS= \dfrac{(23)^{9}-(23)^{8}}{22}$$
$$=\dfrac{23^8(23-1)}{22}$$
$$=\dfrac{23^8\times 22}{22}$$
$$=23^8$$

$$RHS=23^x$$

On comparing $$LHS$$ and $$RHS$$ we get,
$$x=8$$
Question 7
1 / -0

Simplify : $$\displaystyle \frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}$$

A
-4

B
-2

C
-1

D
2

Solution

$$\displaystyle \frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}=\frac{2^{2007}(2^2-1)}{-(2)^{2006}(2^2-1)}$$
$$=-2^{2007-2006}$$
$$=-2$$
Option B is correct.
Question 8
1 / -0

The value of $$5^{1/4}\, \times\, (125)^{0.25}$$ is

A
$$\sqrt 5$$

B
$$5$$

C
$$5\sqrt 5$$

D
$$25$$

Solution

$$5^{1/4} \times (125)^{0.25} = 5^{0.25} \times (5^3)^{0.25}$$

$$= 5^{0.25} \times 5^{(3\times 0.25)}$$

$$ = 5^{0.25} \times 5^{0.75}$$

$$= 5^{(0.25 + 0.75)} = 5^1 = 5$$
Question 9
1 / -0

Simplify: $$\displaystyle \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$

A
$$5^5$$

B
$$5^4$$

C
$$5^2$$

D
$$1$$

Solution

$$ \cfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$

$$=\cfrac { 3^{ -5 }\times (2\times 5)^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times (2\times 3)^{ -5 } } $$

$$=\cfrac { 3^{ -5 }\times 2^{ -5 }\times 5^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times 2^{ -5 }\times 3^{ -5 } } $$

$$=3^{ -5+5 }\times 2^{ -5+5 }\times 5^{ -5+3+7 }$$

$$=5^{5}$$

$$=3125$$
Question 10
1 / -0

$$\displaystyle \left ( \frac{-4}{5} \right )^{4} \times \left ( \frac{-4}{5} \right )^{2} = \frac{16}{25}$$

A
True

B
False

C
Ambiguous

D
Data insffucient

Solution

$$(-\frac { 4 }{ 5 } )^{ 4 }*(-\frac { 4 }{ 5 } )^{ 2 }=(-\frac { 4 }{ 5 } )^{ 4+2 }=(-\frac { 4 }{ 5 } )^{ 6 }\neq \frac { 16 }{ 25 } $$

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Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 4

Result

Logarithm and Antilogarithm Test 4

Score

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SHARING IS CARING

Question 1

$$4$$

$$16$$

$$64$$

$$256.25$$

Solution

Question 2

$$102$$

$$105$$

$$107$$

$$109$$

Solution

Question 3

$$7\, \times\, 8^{-25}$$

$$7\, \times\, 8^{-26}$$

$$8\, \times\, 8^{-26}$$

None of these

Solution

Question 4

$$25$$

$$125$$

$$250$$

$$625$$

Solution

Question 5

$$\displaystyle 2^{69}$$

$$\displaystyle 2^{70}$$

$$\displaystyle 2^{71}$$

$$\displaystyle 2^{72}$$

Solution

Question 6

$$7$$

$$8$$

$$9$$

$$1$$

Solution

Question 7

-4

-2

-1

2

Solution

Question 8

$$\sqrt 5$$

$$5$$

$$5\sqrt 5$$

$$25$$

Solution

Question 9

$$5^5$$

$$5^4$$

$$5^2$$

$$1$$

Solution

Question 10

True

False

Ambiguous

Data insffucient

Solution

User

Question Analysis

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