Mattrab Community - SXC asked a question

Log2aa=x    then, a=(2a) ......(1)

Log3a2a=y    then,2a=(3a)y ......(2)

Log4a 3a=z  then, 3a=(4a)z ......(3)

So, 

a=(2a)x  [from (1)]

Or, a=(3a)xy    [from(2)]

Or, a=(4a)xyz     [from(3)]

Multiplying both sides by 4a,

4a.a=4a.(4a)xyz  

Or,(2a)² =(4a)xyz + 1 

Or,(3a)2y =(4a)xyz+1 

Or,(4a)2yz =(4a)xyz+1 

Or, 2yz = xyz+1 .proved.



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Mattrab Community - SXC asked a question

Here, f: A>B

f(x)= (x-1)/(x+2)  ;  x   -2 

A= {-1,0,1,2,3,4}

B= {-2,1,-1/2,0,1/2,1/4,2/5}

Range = {-2,-1/2,0,1/4,1/2,2/5}

As range is not equal to codomain so the given function is not bijective. 

We can make it bijective by omitting {1} from set B 

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Sushil Bhandari asked a question

Recently we're working to degrade accounts with 0 contributions from creator to learner. If you're a learner and very keen to be a creator, you must keep posting interesting questions and contact to admins from the Facebook Group of Mattrab Community. For being an admin, you must be in grade 12, either completed or recently enrolled, your notes, and all your records and contributions will be verified for that

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Bijaya Rajbhandari asked a question

1) That old man is Aksionov.

2) The speaker is asking about the truth of digging an escape hole .

3) The speaker means the prison's wall.

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