Simplify the expression:
$\left(\frac{e^0 f^2}{f^3 g h^4}\right)^5$
To simplify the expression \(\left(\frac{e^0 f^2}{f^3 g h^4}\right)^5\), we start by analyzing the components inside the parentheses. First, note that any non-zero base raised to the power of zero is 1, so \(e^0 = 1\). This simplifies the numerator inside the expression to \(1 \times f^2 = f^2\).
Next, we simplify the fraction inside the parentheses by combining the terms with base \(f\). Using the quotient rule for exponents, \(\frac{f^a}{f^b} = f^{a-b}\), we calculate \(\frac{f^2}{f^3} = f^{2-3} = f^{-1}\). Substituting this back into our expression, the expression inside the parentheses becomes \(\frac{f^{-1}}{g h^4}\), which can also be written as \(f^{-1} g^{-1} h^{-4}\).
Finally, we apply the power of a product rule, \((x^a y^b z^c)^n = x^{an} y^{bn} z^{cn}\), to distribute the outer exponent of 5 to each term inside the parentheses:
\[(f^{-1} g^{-1} h^{-4})^5 = f^{-1 \times 5} g^{-1 \times 5} h^{-4 \times 5} = f^{-5} g^{-5} h^{-20}.\]
Alternatively, expressed with positive exponents, this is \(\frac{1}{f^5 g^5 h^{20}}\).
In computer science and data storage, exponents are used to describe how data scales. If you have a compression algorithm where a specific unit of data, defined by variables like \(f, g,\) and \(h\), is reduced in size by a certain power, understanding how these exponents combine is crucial for calculating the final storage requirements. By simplifying these expressions, engineers can quickly determine the efficiency of their systems without performing redundant calculations.