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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 :(

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2 Answers 2

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 $$x^{9/30} = 3 \quad \Longleftrightarrow \quad x = 3^{30/9} = 3^{10/3} = \sqrt[3]{3^{10}}$$

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thanks, this last step was the problem. –  toni.a Sep 11 '11 at 21:29
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If $x\gt 0$, then:

  • $\sqrt[3]{x^{-2}} = x^{-2/3}$.
  • $x\sqrt[3]{x^{-2}} = xx^{-2/3} = x^{1/3}$.
  • $\sqrt{x\sqrt[3]{x^{-2}}} = x^{1/6}$.
  • $x^3\sqrt{x\sqrt[3]{x^{-2}}} = x^{19/6}$.
  • $\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}} = x^{19/30}$.
  • $x\sqrt[3]{x} = xx^{1/3} = x^{4/3}$.
  • $\sqrt[4]{x\sqrt[3]{x}} = x^{1/3}$.

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 $x^{1/6}$.

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x must be higher than zero. btw. how do you mean "raise both sides to the 30/9" ? –  toni.a Sep 11 '11 at 21:14
    
@tonia: As $x\gt 0$ a given? If so, good; because the expression actually makes sense for negative $x$s too. As to the latter, I mean that if you have $a=b$, then you also have $a^{30/9} = b^{30/9}$. So take $x^{9/30}=3$, and raise both sides to the $30/9$-th power. That will give you $(x^{9/30})^{30/9} = 3^{30/9}$. Now simplify the left hand side using the rules of exponents. –  Arturo Magidin Sep 11 '11 at 21:18
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