Please, help me to solve this equation:
$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$
I tried to shorten fraction, but I get very weird numbers like $\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}=3,$ and I'm stuck there :(
Please, help me to solve this equation:
$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$
I tried to shorten fraction, but I get very weird numbers like $\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}=3,$ and I'm stuck there :(
You're almost done. Note that $\frac{1}{x} = x^{-1}$, so $\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}} = \sqrt[30]{x^{19}}(\sqrt[3]{x})^{-1} = x^{19/30}x^{-1/3} = x^{9/30}$ So then $x^{9/30} = 3 \quad \Longleftrightarrow \quad x = 3^{30/9} = 3^{10/3} = \sqrt[3]{3^{10}}$
If $x\gt 0$, then:
So the quotient is equal to $x^{19/30}/x^{1/3} = x^{(19/30)-(1/3)} = x^{9/30}$. Your equation is then equivalent to $x^{9/30} = 3$ which can be solved by raising both sides to the $30/9$.
If $x\lt 0$, then you need to throw in a few absolute values, since for example, $\sqrt{x\sqrt[3]{x^{-2}}} = |x|^{1/6}$, instead of $x^{1/6}$.