(a)
x^{-\frac{3}{4}}=\frac{1}{8} , x\ne 0
x^{-\frac{3}{4}}=\frac{1}{2^3}
(x^{\frac{1}{4}})^{-3}=2^{-3}
x^{\frac{1}{4}}=2 \ |^4
x=16
(b)
x^{\frac{4}{3}}=3\sqrt[3]{3}
x^{\frac{4}{3}}=3^1\cdot 3^{\frac{1}{3}}
x^{1\frac{1}{3}}=3^{1\frac{1}{3}}
x=3
c)
x^{-1,5}=3\frac{3}{8} , x\ne 0
x^{-\frac{3}{2}}=\frac{27}{8}
(x^{\frac{1}{2}})^{-3}=(\frac{3}{2})^3
(x^{\frac{1}{2}})^{-3}=(\frac{2}{3})^{-3}
x^{\frac{1}{2}}=\frac{2}{3} \ |^2
x=\frac{4}{9}
(d)
x^{-\frac{3}{2}}=\frac{\sqrt{2}}{4} , x\ne 0
x^{-\frac{3}{2}}=\frac{2^{\frac{1}{2}}}{2^2}
x^{-\frac{3}{2}}=2^{-\frac{3}{2}}
x=2
(e)
x^{\frac{2}{3}}=9
x^{\frac{2}{3}}=3^2 \ |^{\frac{3}{2}}
x=3^3
x=27
(f)
x^{-\frac{5}{6}}=\frac{1}{32} , x\ne 0
(x^{\frac{1}{6}})^{-5}=2^{-5}
x^{\frac{1}{6}}=2 \ |^6
x=64
(g)
\frac{x^{-\frac{4}{3}}}{x^{\frac{1}{6}}}=2 , x\ne 0
x^{-\frac{8}{6}-\frac{1}{6}}=2
x^{-\frac{9}{6}}=2
x^{-\frac{3}{2}}=2 \ |^{-\frac{2}{3}}
x=\frac{1}{2^{\frac{2}{3}}}
x=\frac{1 \cdot 2^{\frac{1}{3}}}{2^{\frac{2}{3}}\cdot 2^{\frac{1}{3}}}
x=\frac{\sqrt[3]2}{2}
(i)
(\frac{2\sqrt{x}}{x^2})^{-3}=[(x\sqrt{x})^{-1}]^{-\frac{1}{2}} , x\ne 0
(\frac{x^2}{2 \cdot x^{\frac{1}{2}}})^3=(x^{\frac{3}{2}})^{\frac{1}{2}}
\frac{x^{\frac{3}{2}}}{2}=\frac{x^{\frac{1}{4}}}{1} proporcja - mnożę “na krzyż”
x^{\frac{3}{2}}=2x^{\frac{1}{4}} \ |^4
x^6=16x\\\\ x^6-16x=0 \\\\ x(x^5-16)=0\\\\ x=0 \ odrzucamy\\\\ lub \\\\x^5=16 \\\\ x=\sqrt[5]{16}