Question

TÌm \(x,y\in N\) biết : a) \(2^{x+1}.3y=12^x\)

b) \(10^x:5^y=20^y\)

c) \(2^x=4^{y-1}\) và \(27y=3^{x+8}\)

Answer 1

a/ \(2^{x+1}.3y=12^x\Leftrightarrow2^x.2.3y=2^x.2^x.3^x\Leftrightarrow2^{x-1}.3^{x-1}=y\Leftrightarrow y=6^{x-1}\)

Vậy \(\left(x;y\right)=\left(x;6^{x-1}\right)\)

b/ \(10^x:5^y=20^y\Leftrightarrow10^x=20^y.5^y\Leftrightarrow10^x=100^y\Leftrightarrow10^x=10^{2y}\Leftrightarrow x=2y\)

Vậy \(\left(x;y\right)=\left(2y;y\right)\)

 

 

Answer 2

c/ Ta có : \(2^x=4^{y-1}\Leftrightarrow2^x=2^{2y-2}\Leftrightarrow x=2y-2\)

Lại có : \(27y=3^{x+8}\Leftrightarrow27y=3^{x+5}.27\Leftrightarrow y=3^{x+5}\)

Suy ra hệ : \(\begin{cases}x=2y-2\\y=3^{x+5}\end{cases}\)