Determine the value of $x$ that satisfies $\sqrt[5]{x\sqrt{x^3}}=3$.
We can first rewrite the term under the fifth root: $x\sqrt{x^3} = x \cdot x^{3/2} = x^{5/2}$. Then we simplify the entire expression on the left side of the equation, which gives  $\sqrt[5]{x^{5/2}}=(x^{5/2})^{1/5} = x ^{(5/2)\cdot(1/5)} = x^{1/2}$. We now have $\sqrt{x}=3$ and we can square each side to find $x=\boxed{9}$.
9
Level 4
Algebra
Problem: Determine the value of $x$ that satisfies $\sqrt[5]{x\sqrt{x^3}}=3$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation $\sqrt[5]{x\sqrt{x^3}}=3$, first simplify the expression inside the fifth root. The term $\sqrt{x^3}$ can be rewritten as $x^{3/2}$. Therefore, the equation becomes $\sqrt[5]{x \cdot x^{3/2}}$. 

Next, combine the exponents of $x$: 
$x \cdot x^{3/2} = x^{1 + 3/2} = x^{5/2}$. 

Now, the equation is $\sqrt[5]{x^{5/2}}=3$. Using the property of exponents, $\sqrt[5]{x^{5/2}}$ can be expressed as $x^{(5/2) \cdot (1/5)} = x^{1/2}$. 

This simplifies the equation to $x^{1/2} = 3$. To eliminate the square root, square both sides: $(x^{1/2})^2 = 3^2$. 

This results in $x = 9$. Thus, the value of $x$ that satisfies the original equation is $9$.<|END_OF_LATENT|> The final answer is $9$.