\sqrt[3]{70 n}(\sqrt[4]{70 n})^{2} For what value of $x$ is the given expression equivalent to $(70 n)^{20 x}$, where $n>1$ ?
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Answer

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Step 1:
: Recognize that the given expression is a cube root and a fourth power multiplied together.

\sqrt[3]{70 n} \times (\sqrt[4]{70 n})^2

Step 2:
: To make the base of both radicals the same, we can rewrite the fourth root as a cube root and a remaining fourth power.

(\sqrt[3]{70 n}) \times (70 n)^{\frac{2}{4}}

Step 3:
: Simplify the expression further.

(\sqrt[3]{70 n}) \times (70 n)^{\frac{1}{2}}

Step 4:
: We want the exponent of $70 n$ to be $20x$.

(\sqrt[3]{70 n}) \times (70 n)^{20x - \frac{1}{3}}

Step 5:
: Since we want the entire expression to be equal to $(70 n)^{20x}$, the exponents on the left side of the equation should be equal to the exponent on the right side of the equation.

\frac{1}{2} = 20x - \frac{1}{3}
Therefore, we have:

Step 6:
: Solve for $x$.

x = \frac{5}{120}

Final Answer

The given expression is equivalent to $(70 n)^{20x}$ when $x = \boxed{\frac{1}{24}}$.