Question

a45 - exponents power rule
score: 9/10 answered: 9/10
question 10
the expression \\(\left(\frac{x^{5}y^{3}z^{2}x^{-4}}{x^{5}y^{4}z^{2}y^{4}}\
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 the numerator and denominator

In the numerator, for \(x\): \(x^{5} \cdot x^{-4}=x^{5 + (-4)} = x^{1}\) (using the rule \(a^{m}\cdot a^{n}=a^{m + n}\)).
In the denominator, for \(x\): \(x^{5}\).

So the fraction inside the parentheses for \(x\) is \(\frac{x^{1}}{x^{5}}=x^{1-5}=x^{-4}\) (using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\)).

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

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

Answer:

The exponent \(r\) of \(x\) is \(\boldsymbol{12}\).

For the exponent \(s\) of \(y\):

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

In the numerator, for \(y\): \(y^{3}\).
In the denominator, for \(y\): \(y^{4}\cdot y^{4}=y^{4 + 4}=y^{8}\) (using \(a^{m}\cdot a^{n}=a^{m + n}\)).

So the fraction inside the parentheses for \(y\) is \(\frac{y^{3}}{y^{8}}=y^{3 - 8}=y^{-5}\) (using \(\frac{a^{m}}{a^{n}}=a^{m - n}\)).

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

Raise \(y^{-5}\) to the power of \(-3\): \((y^{-5})^{-3}=y^{(-5)\times(-3)} = y^{15}\) (using \((a^{m})^{n}=a^{m\times n}\)).

The exponent \(s\) of \(y\) is \(\boldsymbol{15}\).

For the exponent \(t\) of \(z\):

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

In the numerator, for \(z\): \(z^{2}\).
In the denominator, for \(z\): \(z^{2}\).

So the fraction inside the parentheses for \(z\) is \(\frac{z^{2}}{z^{2}}=z^{2 - 2}=z^{0}=1\) (using \(\frac{a^{m}}{a^{n}}=a^{m - n}\), and \(a^{0}=1\) for \(a
eq0\)).

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

Raise \(z^{0}\) to the power of \(-3\): \((z^{0})^{-3}=z^{0\times(-3)} = z^{0}\) (using \((a^{m})^{n}=a^{m\times n}\), and \(0\times(-3)=0\)).

The exponent \(t\) of \(z\) is \(\boldsymbol{0}\).