Question

Tìm \(x;y\) biết 

\(3^x=9^{y-1}\)và \(8^y=2^{x+8}\)

Answer 1

3^x = 9^y-1

=> 3^x = 3^2y-2

=> x = 2y-2

8^y = 2^x+8

=> 2^3y = 2^x+8

=> 3y = x+8

=> 3y = 2y-2+8

=> 3y = 2y+6

=> y = 6

Thay vào, ta có: x = 2y-2

=> x = 2.6 - 2

=> x = 10

KL: x = 10; y = 6

Answer 2

Vì  \(3^x=9^{y-1}\) và \(8^y=2^{x+8}\) nên ta có:

\(3^x.8^y=9^{y-1}.2^{x+8}\)

\(\Rightarrow3^x.2^{3y}=3^{2y}:3^2.2^x.2^8\)

\(\Rightarrow3^x.2^{3y}=3^{2y}.2^x.\frac{256}{9}\)