Question

plz answer this question I will definitely mark you as a brainlist ​

Answer 1

Answer:-

This is the required solution.

Given,

$\sf A = \{ log_{3}(243), log_{5}(125) , log_{2}(64) \}$

First of all, we will evaluate each value,

$\sf log_{3}(243)$

$\sf = log_{3}( {3}^{5} )$

$\sf = 5 log_{3}(3)$

$\sf = 5$

Also,

$\sf log_{5}(125)$

$\sf = log_{5}( {5}^{3} )$

$\sf = 3 log_{5}(5)$

$\sf = 3$

And,

$\sf log_{2}(64) = 6$

So,

$\sf A = \{ 5, 3 , 6\}$

Now,

$\sf B = \{x: x \text{ \: is \: a \: factor \: of \: 20} \}$

So,

$\sf B = \{1, 2, 4, 5, 10 , 20\}$

Now,

$\sf A-B = \{3,6 \}$

i.e, elements present in set A but not in B.

This is the required answer.

Answer 2

Answer:

This is the required solution.

Given,

\sf A = \{ log_{3}(243), log_{5}(125) , log_{2}(64) \}A={log

3

(243),log

5

(125),log

2

(64)}

First of all, we will evaluate each value,

\sf log_{3}(243)log

3

(243)

\sf = log_{3}( {3}^{5} )=log

3

(3

5

)

\sf = 5 log_{3}(3)=5log

3

(3)

\sf = 5=5

Also,

\sf log_{5}(125)log

5

(125)

\sf = log_{5}( {5}^{3} )=log

5

(5

3

)

\sf = 3 log_{5}(5)=3log

5

(5)

\sf = 3=3

And,

\sf log_{2}(64) = 6log

2

(64)=6

So,

\sf A = \{ 5, 3 , 6\}A={5,3,6}

Now,

\sf B = \{x: x \text{ \: is \: a \: factor \: of \: 20} \}B={x:x is a factor of 20}

So,

\sf B = \{1, 2, 4, 5, 10 , 20\}B={1,2,4,5,10,20}

Now,

\sf A-B = \{3,6 \}A−B={3,6}

i.e, elements present in set A but not in B.

This is the required answer.

hope the answer helps u