SAT Question 137cc6fd: Equivalent Expressions answer and explanation with Desmos solution

An expression of the form $\sqrt[k]{a}$, where $k$ is an integer greater than $1$ and $a\ge 0$, is equivalent to $a^{\frac{1}{k}}$. Therefore, the given expression, where $n>1$, is equivalent to ${\left(70n\right)}^{\frac{1}{5}}{\left({\left(70n\right)}^{\frac{1}{6}}\right)}^2$. Applying properties of exponents, this expression can be rewritten as ${\left(70n\right)}^{\frac{1}{5}}{\left(70n\right)}^{\frac{1}{6}\cdot 2}$, or ${\left(70n\right)}^{\frac{1}{5}}{\left(70n\right)}^{\frac{1}{3}}$, which can be rewritten as ${\left(70n\right)}^{\frac{1}{5}+\frac{1}{3}}$, or ${\left(70n\right)}^{\frac{8}{15}}$. It's given that the expression $\sqrt[5]{70n}{\left(\sqrt[6]{70n}\right)}^2$ is equivalent to ${\left(70n\right)}^{30x}$, where $n>1$. It follows that ${\left(70n\right)}^{\frac{8}{15}}$ is equivalent to ${\left(70n\right)}^{30x}$. Therefore, $\frac{8}{15}=30x$. Dividing both sides of this equation by $30$ yields $\frac{8}{450}=x$, or $\frac{4}{225}=x$. Thus, the value of $x$ for which the given expression is equivalent to ${\left(70n\right)}^{30x}$, where $n>1$, is $\frac{4}{225}$. Note that 4/225, .0177, .0178, 0.017, and 0.018 are examples of ways to enter a correct answer.