Math, asked by JayKumar163, 1 year ago

If 9^a+1 = 81^b+2 and (1/3)^3+a = (1/27)^3b. Find the value of a and b.

Answers

Answered by abhay022
9
Hope it will help you !!
Attachments:

JayKumar163: Wrong ans
abhay022: Just calculation mistake .. let me correct it
JayKumar163: Okay sure
abhay022: A= 2/7
abhay022: B = 6/7
JayKumar163: Nope
Answered by rohitkumargupta
22
HELLO DEAR

GIVEN THAT:-


9^(a + 1) = 81^(b + 2)

9^(a + 1) = (9²)^(b + 1)

9^(a + 1) = 9^(2b + 2)

(a + 1) = (2b + 2)

a - 2b = 2 - 1

a - 2b = 1-----------( 1 )



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(1/3)^(3 + a) = (1/27)^(3b)

(1/3)^(3 + a) = [ (1/3)³ ] ^3b

(1/3)^(3 + a) = (1/3)^9b


3 + a = 9b

a - 9b = -3----------( 2 )



from---------(1) & --------(2)


a - 2b = 1
a - 9b = -3
(-)__(+)__(+)
-------------------
7b = 4

b = 4/7 put in ---(1)


we get,


a - 2(4/7) = 1

a = 1 + 8/7

a = (7 + 8)/7

a = 15/7



I HOPE ITS HELP YOU DEAR,
THANKS
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