Question by Summer Clouds

15/11/2017 at 08:32

Bảo Bình 15/11/2017 at 12:09

We have:$2^{18}=4^x\Leftrightarrow2^{18}-4^x=0\Leftrightarrow\left(2^2\right)^9-4^x=0\Leftrightarrow4^9-4^x=0$

$\Leftrightarrow4^z\left(4^{9-x}-1\right)=0$.Because 4^x>0 so 4^9-x=1$\Rightarrow9-x=0\Rightarrow x=9$
Nguyễn Hưng Phát 16/11/2017 at 12:56

We have: $2^{18} = 4^{x} \Leftrightarrow 2^{18} - 4^{x} = 0 \Leftrightarrow {(2^{2})}^{9} - 4^{x} = 0 \Leftrightarrow 4^{9} - 4^{x} = 0$ $\Leftrightarrow 4^{z} (4^{9 - x} - 1) = 0$ .Because 4^x>0 so 4^9-x=1 $\Rightarrow 9 - x = 0 \Rightarrow x = 9$
Phan Thanh Tinh Coordinator 15/11/2017 at 11:33

$2^{18}=\left(2^2\right)^9=4^9$. So, x = 9