Question

if 9^a 1=81^b+2 and (1/3)^3+a=(1/27)^3b, find the values of a an b

Answer 1

Answer:

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

apply this formula not the sum

Step-by-step explanation: