Question

if A and B are natural number such that(1/A)^1/B=0.3bar, then the value of AB is ​

Answer 1

Answer:

AB = (3^B) B

Step-by-step explanation:

Let

z be rational number such that

z = 0.3bar

then

z = 0.3333...                         (1)

Multiplying by 10 on both sides we get

10z = 3.333...                       (2)

Subtracting equation (1) from equation (2) we get

              10z - z = 3.

       ⇒          9z = 3

       ⇒            z = 3 / 9 = 1 / 3          ....(3)

We are given in the question

   z = (1/A)^(1/B)      ...(4)         where A and B are natural numbers

By comparing equations (3) and  (4) we get

(1/A)^(1/B) = 1/3

Taking Bth power on both sides we get

1/A = 1/(3^B)

Taking reciprocal we get

A = 3^B

NOW

AB = (3^B) B.

which is the required result.    

 

Answer 2

Answer:

The value of AB is $B3^B$

Step-by-step explanation:

Given:

$(\frac{1}{A})^\frac{1}{B}=\frac{1}{3}$

Taking logarithm on both sides

$log(\frac{1}{A})^\frac{1}{B}=log\frac{1}{3}$

$\frac{1}{B}log(\frac{1}{A})=log1-log3$

$\frac{1}{B}log(\frac{1}{A})=0-log3$

$\frac{1}{B}log(\frac{1}{A})=-log3$

$log(\frac{1}{A})=-Blog3$

$log(\frac{1}{A})=log3^{-B}$

$log(\frac{1}{A})=log(\frac{1}{3^{B}})$

This implies

$\frac{1}{A}=\frac{1}{3^B}$

$A=3^B$

Then,

$AB=(3^B)B$

$AB=B3^B$

when B=1, AB = 3

when B=2, AB = 18

when B=3, AB = 81

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