Question

a45 - exponents power rule
score: 10/10 answered: 9/10
question 10
the expression \\(\left( \frac{x^4 y^5 z^2 x^{-5}}{x^3 y^2 z^4 y^5} \
ight)^{-3}\\) equals \\(x^r y^s z^t\\)
where \\(r\\), the exponent of \\(x\\), is:
and \\(s\\), the exponent of \\(y\\), is:
and finally \\(t\\), the exponent of \\(z\\), is:
question help: message instructor

Explanation:

Step1: Simplify exponents of \(x\) in numerator and denominator

In the numerator, for \(x\): \(x^{4} \cdot x^{-5}=x^{4 + (-5)} = x^{-1}\) (using \(a^m \cdot a^n=a^{m + n}\)).
In the denominator, for \(x\): \(x^{3}\).
So the fraction inside the parentheses for \(x\) is \(\frac{x^{-1}}{x^{3}}=x^{-1-3}=x^{-4}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)).

Step2: Apply the power of \(-3\) to the exponent of \(x\)

Now, raise this to the power of \(-3\): \((x^{-4})^{-3}=x^{(-4)\times(-3)} = x^{12}\) (using \((a^m)^n=a^{m\times n}\)). So \(r = 12\).

Step3: Simplify exponents of \(y\) in numerator and denominator

In the numerator, for \(y\): \(y^{5}\).
In the denominator, for \(y\): \(y^{2}\cdot y^{5}=y^{2 + 5}=y^{7}\) (using \(a^m \cdot a^n=a^{m + n}\)).
So the fraction inside the parentheses for \(y\) is \(\frac{y^{5}}{y^{7}}=y^{5 - 7}=y^{-2}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)).

Step4: Apply the power of \(-3\) to the exponent of \(y\)

Raise to the power of \(-3\): \((y^{-2})^{-3}=y^{(-2)\times(-3)} = y^{6}\) (using \((a^m)^n=a^{m\times n}\)). So \(s = 6\).

Step5: Simplify exponents of \(z\) in numerator and denominator

In the numerator, for \(z\): \(z^{2}\).
In the denominator, for \(z\): \(z^{4}\).
So the fraction inside the parentheses for \(z\) is \(\frac{z^{2}}{z^{4}}=z^{2 - 4}=z^{-2}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)).

Step6: Apply the power of \(-3\) to the exponent of \(z\)

Raise to the power of \(-3\): \((z^{-2})^{-3}=z^{(-2)\times(-3)} = z^{6}\) (using \((a^m)^n=a^{m\times n}\)). So \(t = 6\).

Answer:

\(r = 12\), \(s = 6\), \(t = 6\)