b. \(10^x:5^y=20^y\)
\(\Leftrightarrow10^x=20^y.5^y=100^y\)
\(\Leftrightarrow10^x=\left(10^2\right)^y\)
\(\Leftrightarrow x=2y\)
a, 2x+1 . 3y = 12x
=> \(3^y=\dfrac{12^x}{2^{x+1}}\)
=> \(3^y=\dfrac{2^x.3^x.2^x}{2^{x+1}}\)
=> \(3^y=\dfrac{2^{x+1}.3^x}{2^{x+1}}\)
=> 3y=3x
=> x=y
a. Ta có: \(2^{x+1}.3^y=12^x\)
\(\Leftrightarrow2^{x+1}.3^y=\left(2^2.3\right)^x\)
\(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}2^{x+1}=2^{2x}\\3^y=3^x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\y=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=x\end{matrix}\right.\)
Vậy x = y = 1
b. 10x:5y=20y
⇔10x=20y.5y=100y
⇔10x=(102)y
⇔x=2y
