Question

Consider the system of equations

3^y-1=2^x

3^y+2^{4-x}=11

If the solutions of the system are of the form (x_1,y_1),(x_2,y_2) what is x_1+x_2+y_1+y_2​

Answer 1

Given :   system of equations $3^y-1=2^x$

$3^y+2^{4-x}=11$

(x₁ , y₁) & (x₂ , y₂)  solutions of the system

To Find : x₁ + x₂ + y₁ + y₂

Solution:

$3^y-1=2^x$

$3^{y}= 2^x+1$

$3^y+2^{4-x}=11$

$2^x+1+2^{4-x}=11$

=> 2ˣ  -10 + 2⁴/2ˣ  = 0

=>  (2ˣ)²  -10* 2ˣ + 16  = 0

=>  (2ˣ)²  -8* 2ˣ  - 2 ** 2ˣ + 16  = 0

=> (2ˣ - 8)(2ˣ - 2) = 0

=> 2ˣ - 8 = 0

=> 2ˣ = 2³  => x = 3

2ˣ - 2  = 0  => x = 1

$3^{y}= 2^1+1= 3=3^1$

=> y = 1

$3^{y}= 2^3+1= 9=3^2$

=> y = 2

x₁= 1  , y₁ = 1

x₂ = 3  , y₂ = 2

1 + 3 + 1 + 2  =  7

x₁ + x₂ + y₁ + y₂  = 7

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