Нужна с логарифмами решите систему уравнений

Самина2006 Самина2006    3   13.06.2019 23:30    2

Ответы
Azamatovna11 Azamatovna11  10.07.2020 19:46
\left \{ {{log_{2}x-log_{2^{2}}y=0} \atop {log_{2^{2}}x+log_{2}y=1}} \right.

\left \{ {{log_{2}x-0.5*log_{2}y=0} \atop {0.5*log_{2}x+log_{2}y=1}} \right.

\left \{ {{log_{2}x-log_{2}(\sqrt{y})=0} \atop {log_{2}(\sqrt{x})+log_{2}y=1}} \right.

\left \{ {{log_{2}(\frac{x}{\sqrt{y}})=log_{2}1} \atop {log_{2}(\sqrt{x}*y)=log_{2}2}} \right.

\left \{ {{\frac{x}{\sqrt{y}}=1} \atop {\sqrt{x}*y=2} \right.

\left \{ {{x=\sqrt{y}} \atop {\sqrt{x}*y=2} \right.

\left \{ {{x^{2}=y} \atop {\sqrt{x}*x^{2}=2} \right.

\left \{ {{x^{2}=y} \atop {x^{\frac{5}{2}}=2} \right.

\left \{ {{x^{2}=y} \atop {x=2^{\frac{2}{5}}} \right.

\left \{ {{y=(2^{\frac{2}{5}})^{2}=2^{\frac{4}{5}}} \atop {x=2^{\frac{2}{5}}} \right.

ответ: (2²/⁵; 2⁴/⁵)
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