Question

properties of exponents

1. ((4x^{3}y)^{0})
2. ((5x^{4})^{-3})
3. (x^{13} cdot x^{-6})
4. ((5x^{4})^{-3})

Explanation:

Problem 38: \((4x^{3}y)^{0}\)

Step 1: Recall the zero - exponent property

The zero - exponent property states that for any non - zero real number \(a\), \(a^{0}=1\). In the expression \((4x^{3}y)^{0}\), as long as \(4x^{3}y
eq0\) (which is true for all real numbers \(x\) and \(y\) except when \(x = 0\) and \(y = 0\) simultaneously, but in the context of exponent rules, we assume the base is non - zero), we can apply the zero - exponent property.
So, \((4x^{3}y)^{0}=1\)

Step 1: Recall the power of a product rule and negative exponent rule

The power of a product rule is \((ab)^{n}=a^{n}b^{n}\), and the negative exponent rule is \(a^{-n}=\frac{1}{a^{n}}\) (\(a
eq0\)).
First, apply the power of a product rule to \((5x^{4})^{-3}\):
\((5x^{4})^{-3}=5^{-3}\times(x^{4})^{-3}\)

Step 2: Simplify the exponents

For \(5^{-3}\), using the negative exponent rule, \(5^{-3}=\frac{1}{5^{3}}=\frac{1}{125}\)
For \((x^{4})^{-3}\), use the power of a power rule \((a^{m})^{n}=a^{mn}\), so \((x^{4})^{-3}=x^{4\times(-3)}=x^{-12}=\frac{1}{x^{12}}\) (by the negative exponent rule)
Then, multiply the two results: \(5^{-3}\times(x^{4})^{-3}=\frac{1}{125}\times\frac{1}{x^{12}}=\frac{1}{125x^{12}}\)

Step 1: Recall the product of powers rule

The product of powers rule states that for any non - zero real number \(a\) and integers \(m\) and \(n\), \(a^{m}\cdot a^{n}=a^{m + n}\)
Applying this rule to \(x^{13}\cdot x^{-6}\), we get:
\(x^{13}\cdot x^{-6}=x^{13+(-6)}\)

Step 2: Simplify the exponent

\(13+(-6)=7\), so \(x^{13+(-6)}=x^{7}\)

Answer:

\(1\)

Problem 39: \((5x^{4})^{-3}\)