Question

simplify. express your answer using positive exponents.\\(\frac{r^{7}s^{-8}t^{0}\cdot r^{6}s^{0}t^{0}}{r^{4}s^{0}t}\\)

Explanation:

Step1: Use exponent rules for multiplication (same base, add exponents) in numerator

For \( r \): \( r^7 \cdot r^6 = r^{7 + 6}=r^{13} \)
For \( s \): \( s^{-8} \cdot s^0 = s^{-8 + 0}=s^{-8} \)
For \( t \): \( t^0 \cdot t^0 = t^{0 + 0}=t^0 = 1 \) (since \( a^0 = 1 \) for \( a
eq 0 \))
So numerator becomes \( r^{13}s^{-8} \cdot 1 = r^{13}s^{-8} \)

Step2: Divide by denominator using exponent rule (same base, subtract exponents)

For \( r \): \( r^{13} \div r^4 = r^{13 - 4}=r^9 \)
For \( s \): \( s^{-8} \div s^0 = s^{-8 - 0}=s^{-8} \) (since \( s^0 = 1 \))
For \( t \): \( 1 \div t = t^{0 - 1}=t^{-1}=\frac{1}{t} \) (but we need positive exponents, so handle \( s^{-8} \) and \( t^{-1} \))

Step3: Convert negative exponents to positive

\( s^{-8}=\frac{1}{s^8} \), so \( s^{-8} \) in the expression becomes \( \frac{1}{s^8} \), and \( t^{-1}=\frac{1}{t} \). But wait, let's re - check the \( t \) terms: original numerator \( t \) terms were \( t^0 \cdot t^0 = 1 \), denominator \( t \) is \( t^1 \), so \( t^{0}\div t^{1}=t^{-1}=\frac{1}{t} \). However, let's re - do the division step more accurately:

After numerator is \( r^{13}s^{-8}t^{0} \) (since \( t^0 \cdot t^0 = t^0 \)) and denominator is \( r^4s^0t^1 \).

So:

\( r \): \( 13 - 4 = 9 \)

\( s \): \( - 8-0=-8 \)

\( t \): \( 0 - 1=-1 \)

Now, rewrite with positive exponents:

\( r^{9}\cdot\frac{1}{s^{8}}\cdot\frac{1}{t}=\frac{r^{9}}{s^{8}t} \)

Wait, but let's check the \( t \) in denominator again. Wait, the original numerator's \( t \) is \( t^0 \cdot t^0 = t^0 \), denominator's \( t \) is \( t^1 \), so \( t^{0}\div t^{1}=t^{-1}=\frac{1}{t} \). And \( s^{-8}=\frac{1}{s^{8}} \). So combining:

\( r^{9}\times\frac{1}{s^{8}}\times\frac{1}{t}=\frac{r^{9}}{s^{8}t} \)

Wait, but let's verify the exponent rules again:

• Product rule: \( a^m \cdot a^n=a^{m + n} \)
• Quotient rule: \( a^m\div a^n=a^{m - n} \)
• Zero exponent: \( a^0 = 1,a

eq0 \)

• Negative exponent: \( a^{-n}=\frac{1}{a^n} \)

Let's re - compute the entire expression:

$$
LATEXBLOCK0
$$

Answer:

\(\frac{r^{9}}{s^{8}t}\)