Question

Решите уравнения б)(1/64)^x=√1/2 в)2^x2-8x+19=16 г)lg x-4/2-x=0 нужна

Answer 1

Б)$2^{-6x} = 2^{-1/2}$
-6x=-1/2
x=1/12
в)$2^{x^2-8x+19} =2^4$
x²-8x+19=4
x²-8x+15=0
x1+x2=8 U x1*x2=15
x1=3 U x2=5
г)(x-4)/(2-x)>0
x=4 U x=2
x∈(2;4)
(x-4)/(2-x)=1
x-4=2-x
x+x=2+4
2x=6
x=3

Answer 2

$b)\ (\frac{1}{64})^x=\sqrt{\frac{1}{2}}\\ ((\frac{1}{\sqrt2})^{12})^x=\frac{1}{\sqrt2}\\ (\frac{1}{\sqrt2})^{12x}=\frac{1}{\sqrt2}\\ 12x=1\\ x=\frac{1}{12}\\ v)\ 2^{x^2-8x+19}=16\\ 2^{x^2-8x+19}=2^4\\ x^2-8x+19=4\\ x^2-8x+15=0\\ |x=3\\ |x=5\\ g)\ \frac{lgx-4}{2-x}=0\\ x0;\ x\neq2\\ lgx-4=0\\ lgx=4\\ lgx=lg10^4\\ x=10^4=10000\\$