Question

The number of solutions of $\sf{\:3^{x+y}=243\:}$ and $\sf{\:243^{x-y}=3\:}$
(a) 1
(b) 9
(c) 4
(d) infinite ​

Answer 1

Answer: (a) 1

Explanation:

3^(x + y) = 243

=> 3^(x + y) = 3^5

∴1 x + 1 y - 5 = 0 (bases are same)

& 243^(x - y) = 3

=> (3^1/5)^(x - y)= 3^1

=> 3^([1/5(x - y))] = 3^1

∴ 1/5(x - y) = 1 (bases are same)

=> 1 x - 1 y - 5 = 0.

Now, comparing the ratios of the coefficients of x, y and constants :-

1/1 ≠ 1/(- 1) = (-5)/(-5)

i.e, a1/a2 ≠ b1/b2.

So, it is a consistent pair & has a unique solution.

Answer 2

Answer:

(a) 1

Step-by-step explanation:

3^(x + y) = 243

=> 3^(x + y) = 3^5

-1x+1y-5=0 (bases are same)

& 243^(x - y) = 3

=> (3^1/5)^(x - y)= 3^1

=> 3^([1/5(x - y))] = 3^1

.. 1/5(x - y) = 1 (bases are same)

=> 1x-1y- 5 = 0.

Now, comparing the ratios of the

coefficients of x, y and constants :

1/1 1/(-1) = (-5)/(-5)

i.e, a1/a2b1/b2.

So, it is a consistent pair & has a unique

solution.