Question

SOLVE :

3^x + 3^y = 4 ; 3^-x + 3^-y = 4/3 ​

Answer 1

Answer:

x = 1 and y = 0 or x = 0 and y = 1

Step-by-step explanation:

Given a pair of equations such that,

${3}^{x} + {3}^{y} = 4$

And

${3}^{ - x} + {3}^{ - y} = \frac{4}{3}$

Let's assume that,

3^x = u and 3^y = v

Therefore, we will get,

$u + v = 4$

And,

$\frac{1}{u} + \frac{1}{v} = \frac{4}{3}$

Solving, further, we will get,

$= > \frac{u + v}{uv} = \frac{4}{3}$

Substituting the value, we get,

$= > \frac{4}{uv} = \frac{4}{3} \\ \\ = > \frac{1}{uv} = \frac{1}{3} \\ \\ = > uv = 3$

Substituting again, we get,

$= > {3}^{x} \times {3}^{y} = 3 \\ \\ = > {3}^{(x + y)} = {3}^{1} \\ \\ = > x + y = 1$

Now, we have,

${3}^{x} + {3}^{y} = 4$

Therefore, we will get,

$= > {3}^{x} + {3}^{y} = 3 + 1 \\ \\ = > {3}^{ x } + {3}^{y} = {3}^{1} + {3}^{0}$

On comparing both sides, we get,

x = 1 and y = 0 or x = 0 and y = 1

Answer 2

Answer:

hope these answer may help you