Logarithm and Antilogarithm Test 1

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Logarithm and Antilogarithm Test 1

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$7^{\frac{1}{2}}. 8^{\frac{1}{2}}$$ is :

A
$$28^{\frac{1}{2}}$$

B
$$56^{\frac{1}{2}}$$

C
$$14^{\frac{1}{2}}$$

D
$$42^{\frac{1}{2}}$$

Solution

$$\Rightarrow$$ $$(7)^{\tfrac{1}{2}}.(8)^{\frac{1}{2}}$$

$$\Rightarrow$$ $$(7\times 8)^{\frac{1}{2}}$$ $$[\,\because\,m^x.n^x=(m\times n)^x]$$

$$\Rightarrow$$ $$56^{\frac{1}{2}}$$
Question 2
1 / -0

Find the product. $$(a^2) (2a^{22}) (4a^{26})$$

A
$$8a^{40}$$

B
$$8a^{50}$$

C
$$8a^{30}$$

D
$$8^{20a}$$

Solution

$$(a^2) (2a^{22}) (4a^{26})$$
$$=(a^2)\times (2a^{22})\times (4a^{26})$$
$$=8a^{2+22+26}$$
$$=8a^{50}$$
Question 3
1 / -0

$$\sqrt [4]{\sqrt [3]{2^{2}}}$$ equal

A
$$2^{\dfrac {-1}{6}}$$

B
$$2^{-6}$$

C
$$2^{\dfrac {1}{6}}$$

D
$$2^{6}$$

Solution

$$\Rightarrow$$ $$\sqrt[4]{\sqrt[3]{2^2}}$$

$$\Rightarrow$$ $$[(2^2)^\tfrac{1}{3}]^{\tfrac{1}{4}}$$

$$\Rightarrow$$ $$(2)^{2\times \tfrac{1}{3}\times \tfrac{1}{4}}$$

$$\Rightarrow$$ $$2^{\tfrac{1}{6}}$$

$$\therefore$$ $$\sqrt[4]{\sqrt[3]{2^2}}=2^{\tfrac{1}{6}}$$
Question 4
1 / -0

If $$ \log 2 = 0.3010$$ and $$ \log 3 = 0.4771$$, then the value of $$\log 6$$ will be

A
$$0.7781$$

B
$$0.6991$$

C
$$0.8066$$

D
$$0.6981$$

Solution

Using the Logarithmic property,

$$\log ab = \log a+ \log b$$

$$ \log 6= \log(2\times3) $$

$$= \log 2+ \log 3$$

$$=0.3010+0.4771$$

$$=0.7781$$
Question 5
1 / -0

The value of $$\cfrac{3^0+7^0}{5^0}$$ is:

A
$$2$$

B
$$0$$

C
$$\dfrac{9}{5}$$

D
$$\dfrac{1}{5}$$

Solution

$$\Rightarrow$$ $$\dfrac{3^0+7^0}{5^0}$$

$$\Rightarrow$$ $$\dfrac{1+1}{1}$$ $$[\,\because\,x^0=1]$$

$$\Rightarrow$$ $$\dfrac{2}{1}$$

$$\Rightarrow$$ $$2$$

$$\therefore$$ $$\dfrac{3^0+7^0}{5^0}=2$$
Question 6
1 / -0

Simplified value of $$(25)^{\frac{1}{3}} \times (5)^{\frac{1}{3}}$$ is :

A
$$25$$

B
$$3$$

C
$$1$$

D
$$5$$

Solution

$$(25)^{\tfrac{1}{3}}\times (5)^{\tfrac{1}{3}}$$ $$\Rightarrow$$ $$(25\times 5)^\tfrac{1}{3}$$ ...$$[\because\,m^x\times n^x=(m\times n)^x]$$

$$\Rightarrow$$ $$(125)^\tfrac{1}{3}$$

$$\Rightarrow$$ $$\sqrt[3]{125}=5$$

$$\therefore$$ $$(25)^{\tfrac{1}{3}}\times (5)^{\tfrac{1}{3}}=5$$
Question 7
1 / -0

If $$\displaystyle \log_3 x = 0$$, then value of $$x$$ is equal to

A
$$2$$

B
$$4$$

C
$$1$$

D
$$3$$

Solution

Given, $$\log_3x = 0$$
$$\implies x=3^0$$ (Since we know that $$\log_ab=t\Rightarrow b=a^t$$)
$$\therefore x=1$$.
Question 8
1 / -0

Given that $$4^{n+1} = 256$$, find the value of $$n$$.

A
$$2$$

B
$$3$$

C
$$5$$

D
$$63$$

Solution

$$4^{n+1}=256$$
Taking log on both sides to the base 2
$$\Rightarrow\: \log_{2}{4^{n+1}}=\log_{2}{256}$$
$$\Rightarrow\:(n+1)\log_{2}{4}=\log_{2}{2^{8}}$$
$$\Rightarrow\:(n+1).\log_{2}{2^2}=8$$
$$\Rightarrow\:(n+1).2=8$$
$$\Rightarrow\:(n+1)=4$$
$$\Rightarrow\:n=3$$
Question 9
1 / -0

The value of $$\displaystyle \log_5 1$$ is

A
$$0$$

B
$$1$$

C
$$3$$

D
$$5$$

Solution

Consider, $$ \log _{ 5 }{ 1 } =x$$
$$\implies 1 = { 5 }^{ x }$$
$$\implies 5^0 = 5^x$$
$$\implies x = 0 $$.
Question 10
1 / -0

The exponential form of $$\log_{10}1 = 0$$ is $$10^{m} = 1$$, then the value of $$m$$ is

A
$$2$$

B
$$0$$

C
$$1$$

D
$$6$$

Solution

$$ \log _{ 10 }{ 1 }= 0\\ 1 = { 10 }^{ 0 }$$

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Re-Attempt Weekly Quiz Competition

Logarithm and Antilogarithm Test 1

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Logarithm and Antilogarithm Test 1

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SHARING IS CARING

Question 1

$$28^{\frac{1}{2}}$$

$$56^{\frac{1}{2}}$$

$$14^{\frac{1}{2}}$$

$$42^{\frac{1}{2}}$$

Solution

Question 2

$$8a^{40}$$

$$8a^{50}$$

$$8a^{30}$$

$$8^{20a}$$

Solution

Question 3

$$2^{\dfrac {-1}{6}}$$

$$2^{-6}$$

$$2^{\dfrac {1}{6}}$$

$$2^{6}$$

Solution

Question 4

$$0.7781$$

$$0.6991$$

$$0.8066$$

$$0.6981$$

Solution

Question 5

$$2$$

$$0$$

$$\dfrac{9}{5}$$

$$\dfrac{1}{5}$$

Solution

Question 6

$$25$$

$$3$$

$$1$$

$$5$$

Solution

Question 7

$$2$$

$$4$$

$$1$$

$$3$$

Solution

Question 8

$$2$$

$$3$$

$$5$$

$$63$$

Solution

Question 9

$$0$$

$$1$$

$$3$$

$$5$$

Solution

Question 10

$$2$$

$$0$$

$$1$$

$$6$$

Solution

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