- First, recognize the zero exponent rule: any non-zero number raised to the power of zero is 1, so \(\left(\frac{3}{5}\right)^0 = 1\).
- Substitute this into the expression, which results in \([1]^{-2}\).
- Apply the property that 1 raised to any power is 1, giving \([1]^{-2} = 1\).
- Final Answer: \(\boxed{1}\).
Explanation
- Analyzing the Inner Expression
To evaluate the expression \(\left[\left(\frac{3}{5}\right)^0\right]^{-2}\), we analyze its components using fundamental exponent properties. The expression consists of a base \(\frac{3}{5}\) raised to the power of \(0\), and the entire result subsequently raised to the power of \(-2\). We recall that any non-zero real number raised to the zero power is 1. Thus, \(\left(\frac{3}{5}\right)^0 = 1\).
- Simplifying the Final Expression
With the inner part evaluated, the expression simplifies to \([1]^{-2}\). We now apply the negative exponent rule, which states that for any non-zero base \(a\), \(a^{-n} = \frac{1}{a^n}\). In this case, \(a = 1\) and \(n = 2\), so we have:
\[[1]^{-2} = \frac{1}{1^2}\]
Since \(1^2 = 1 \times 1 = 1\), the expression becomes \(\frac{1}{1} = 1\). Alternatively, we know that 1 raised to any power is always 1, because 1 multiplied by itself any number of times remains 1.
- Final Result
Based on the steps above, the evaluation of the expression \(\left[\left(\frac{3}{5}\right)^0\right]^{-2}\) yields 1. Therefore, the value of the expression is \(\boxed{1}\).
Examples
Exponent rules, such as the zero exponent property, are essential in various fields like finance and science. For instance, in compound interest calculations, a principal amount at time \(t=0\) is essentially the initial investment, which can be modeled as \(P \times (1+r)^0 = P\). Similarly, this property helps in standardizing scientific notation and simplifying complex physical formulas. Understanding how exponents interact allows for more efficient calculations and cleaner mathematical models in everyday professional contexts.