Question

Помогите пожалуйста

Answer 1

$5)\; \; 5^{2x}-4\cdot 5^{x}=5\\\\t=5^{x}>0\; \; ,\; \; \; t^2-4t-5=0\; ,\; \; t_1=-1<0\; ,\; t_2=5>0\\\\5^{x}=5\; \; \Rightarrow \; \; \underline {x=1}\\\\\\6)\; \; (\frac{1}{64})^{x}=\sqrt{\frac{1}{8}}\\\\(\frac{1}{8})^{2x}=(\frac{1}{8})^{1/2}\; \; ,\; \; 2x=\frac{1}{2}\; \; ,\; \; \underline {x=\frac{1}{4}}$

$7)\; \; (\frac{1}{4})^{4x-10}=\frac{1}{16}\\\\(\frac{1}{4})^{4x-10}=(\frac{1}{4})^2\; \; ,\; \; 4x-10=2\; \; ,\; \; 4x=12\; ,\; \; \underline {x=3}\\\\\\8)\; \; 9^{x-10}=\frac{1}{3}\\\\3^{2(x-10)}=3^{-1}\; \; ,\; \; 2x-20=-1\; \; ,\; \; 2x=19\; \; ,\; \; \underline {x=9,5}$

Answer 2

$5)5^{2x}-4*5^{x}=5\\\\5^{x} =m,m>0\\\\m^{2}-4m-5=0\\\\m_{1}=5\\\\m_{2}=-1-neyd\\\\5^{x} =5\\\\x=1\\\\Otvet:\boxed{1}$

$6)(\frac{1}{64})^{x}=\sqrt{\frac{1}{8}}\\\\(\frac{1}{8})^{2x}=(\frac{1}{8})^{\frac{1}{2}}\\\\2x=\frac{1}{2}\\\\x=0,25\\\\Otvet:\boxed{0,25}$

$7)(\frac{1}{4})^{4x-10}=\frac{1}{16}\\\\(\frac{1}{4})^{4x-10}=(\frac{1}{4})^{2}\\\\4x-10=2\\\\4x=12\\\\x=3\\\\Otvet.\boxed{3}$

$8)9^{x-10}=\frac{1}{3}\\\\3^{2x-20}=3^{-1}\\\\2x-20=-1\\\\2x=19\\\\x=9,5\\\\Otvet:\boxed{9,5}$