Question

1. (2; 2,5)
2.
3.(2; корень из 10)
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Answer 1

ОДЗ : x > 0

$6\log_{8}^{2}x-5\log_{8}x+1=0\\\\\log_{8}x=m\\\\6m^{2}-5m+1=0\\\\D=(-5)^{2}-4\cdot6\cdot1=25-24=1\\\\m_{1} =\frac{5-1}{12} =\frac{1}{3} \\\\m_{2}=\frac{5+1}{12} =\frac{1}{2}\\\\1)\log_{8} x=\frac{1}{3} \\\\x_{1}=8^{\frac{1}{3 }}=\sqrt[3]{8} =2 \\\\2)\log_{8} x=\frac{1}{2}\\\\x_{2}=8^{\frac{1}{2} }=\sqrt{8}=2\sqrt{2}\\\\Otvet:\boxed{2\sqrt{2}}$

$2)4^{x}-2^{x+3} +15=0\\\\(2^{x} )^{2}-2^{x}\cdot2^3+15=0\\\\(2^{x} )^{2}-2^{x}\cdot8 +15=0\\\\2^{x} =m \ , \ m0\\\\m^{2} -8m+15=0\\\\D=(-8)^{2} -4\cdot15=64-60=4=2^{2}\\\\m_{1} =\frac{8-2}{2}=3\\\\m_{2}=\frac{8+2}{2}=5\\\\1)2^{x} =3\\\\x_{1} =\log_{2}3\\\\2)2^{x}=5\\\\x_{2}=\log_{2}5 \\\\Otvet:\boxed{x=\log_{2}5}$