Algebra: A Combined Approach (4th Edition)

$80\pi x^{5}$
We are given that the silo has a radius of $4x$ meters and a height of $5x^{3}$ meters. We know that the volume of a cylinder can be calculated as $V=\pi r^{2}h$. We can plug in the given values and solve. $V=\pi\times(4x)^{2}\times5x^{3}$ Based on the power of a product rule, we know that $(ab)^{n}=a^{n}b^{n}$ (where $n$ is a positive integer and $a$ and $b$ are real numbers). Therefore, $\pi\times(4x)^{2}\times5x^{3}=\pi\times4^{2}\times x^{2}\times5x^{3}=\pi\times16\times x^{2}\times5x^{3}$. Based on the product rule for exponents, we know that $a^{m}\times a^{n}=a^{m+n}$ (where $m$ and $n$ are positive integers and $a$ is a real number). Therefore, $\pi\times16\times x^{2}\times5x^{3}=(16\times5\times\pi)\times (x^{2}\times x^{3})=80\pi\times x^{2+3}=80\pi x^{5}$.