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Chapter 11: Problem 61

Solve the given problems.Is it true that \(\left[-2^{0}-(-1)^{0}\right]^{0}=1 ?\)

Short Answer

Expert verified
Yes, the given expression simplifies to 1.

Step by step solution

01

Evaluate the Exponents in the Inner Expression

Start by evaluating the exponents within the brackets. The expression is given as \(-2^{0} - (-1)^{0}\). According to the rule of exponents, any non-zero number raised to the power of 0 is 1. Thus, \(2^{0}\) is 1 and \((-1)^{0}\) is also 1.
02

Resolve the Inner Bracket Expression

Substitute the values obtained from Step 1 into the expression: \(-2^{0} - (-1)^{0} = -1 - 1\). Simplify this to obtain \(-1 - 1 = -2\).
03

Evaluate the Outer Exponent

Finally, raise the resulting value from the inner bracket to the power 0, according to the original expression \([-2]^{0}\). Any number except 0 raised to the power 0 is 1. Hence, \([-2]^{0} = 1\).
04

Confirm the Original Expression

Combining all steps, \([-2^{0}-(-1)^{0}]^{0} = 1\), which confirms that the original expression simplifies to 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Power of Zero Rule
The power of zero rule is a fundamental principle in mathematics, particularly in exponentiation. When any non-zero number is raised to the power of zero, the result is always 1. This rule helps simplify expressions that involve exponents. Such as in our exercise where both \(2^{0}\) and \((-1)^{0}\) are evaluated to 1.

It's vital to remember that the base number must be non-zero since \(0^{0}\) is usually considered indeterminate in most contexts.

The simplicity of this rule plays a crucial role in various calculations, allowing complex expressions to reduce significantly. This approach allows us to handle equations more efficiently during the solving process.
The Importance of Inner Bracket Evaluation
Inner bracket evaluation refers to the practice of resolving expressions within parentheses first, a key step in order of operations (often remembered using PEMDAS/BODMAS).

In the context of our problem, before addressing the exponent outside, you need to deal with the expression inside the brackets: \(-2^{0} - (-1)^{0}\).

By evaluating each of these exponents using the power of zero rule, we substitute them into the expression to find \(-1 - 1\).

This simplifies the expression inside the brackets to \(-2\), setting the stage for applying other rules such as the outer exponentiation next.

Mastering inner bracket evaluation is essential for solving equations step-by-step, ensuring each part is accurately handled.
Exploring Exponent Simplification
Once inner bracket evaluation is complete and any required calculations within the parentheses are done, exponent simplification plays a crucial role, particularly when raising expressions to a power.

In this exercise, after evaluating the inner bracket to get \(-2\), we must consider the outer exponent.

Raising \(-2\) to the power of zero simplifies to 1, according to the power of zero rule discussed earlier.

Exponent simplification ensures that even complex expressions can be understood and reduced systematically.

By taking each part of an expression, simplifying with known rules, and methodically applying these, students can solve equations smoothly.

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Most popular questions from this chapter

In Exercises \(53-76,\) solve the given problems. Solve for \(x: 2^{5 x}=2^{7}\left(2^{2 x}\right)^{2}\)

Solve the given problems.In optics, the combined focal length \(F\) of two lenses is given by \(F=\left[f_{1}^{-1}+f_{2}^{-1}+d\left(f_{1} f_{2}\right)^{-1}\right]^{-1},\) where \(f_{1}\) and \(f_{2}\) are the focal lengths of the lenses and \(d\) is the distance between them. Simplify the right side of this equation.

In Exercises \(53-76,\) solve the given problems. When studying a solar energy system, the units encountered are \(\mathrm{kg} \cdot \mathrm{s}^{-1}\left(\mathrm{m} \cdot \mathrm{s}^{-2}\right)^{2} .\) Simplify these units and include joules (see Example 4 ) and only positive exponents in the final result.

Perform the indicated operations. The withdrawal resistance \(R\) of a nail of diameter \(d\) indicates its holding power. One formula for \(R\) is \(R=k s^{5 / 2} d h,\) where \(k\) is a constant, \(s\) is the specific gravity of the wood, and \(h\) is the depth of the nail in the wood. Solve for \(s\) using fractional exponents in the result.

In Exercises \(5-52,\) express each of the given expressions in simplest form with only positive exponents. $$\left(\frac{3 a^{2}}{4 b}\right)^{-3}\left(\frac{4}{a}\right)^{-5}$$
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