Question

Tìmm x, y

5^x+2= 25^y và 27^y = 81.3^x+4

Answer 1

\(\left\{{}\begin{matrix}5^{x+2}=25^y\\27^y=81\cdot3^{x+4}\end{matrix}\right.\)

Từ \(27^y=81\cdot3^{x+4}\Rightarrow\left(3^3\right)^y=3^4\cdot3^{x+4}\)

\(\Rightarrow3^{3y}=3^{x+8}\)\(\Rightarrow3y=x+8\left(1\right)\)

Lại có: \(5^{x+2}=25^y\Rightarrow5^{x+2}=\left(5^2\right)^y\)

\(\Rightarrow5^{x+2}=5^{2y}\Rightarrow x+2=2y\left(2\right)\)

Thay \(\left(2\right)\) vào \(\left(1\right)\) ta có:

\(\left(1\right)\Leftrightarrow3y=2y+6\Leftrightarrow y=6\)

\(\Rightarrow x+2=2y=2\cdot6=12\Rightarrow x=10\)