Question

express the following fraction in simplest form, only using positive exponents.
\\(\frac{-2r^{-3}}{2(r^{-3})^5}\\)
answer attempt 2 out of 2
\\(-1r^{-18}\\)

Explanation:

Step1: Simplify coefficients

The coefficient of the numerator is -2 and the coefficient of the denominator is 2. Divide them: $\frac{-2}{2} = -1$.

Step2: Simplify the variable part using exponent rules

For the variable $r$, we use the rule $(a^m)^n = a^{mn}$ for the denominator: $(r^{-3})^5 = r^{-3\times5}=r^{-15}$.
Then we use the rule $\frac{a^m}{a^n}=a^{m - n}$ for the fraction of the variable parts: $\frac{r^{-3}}{r^{-15}} = r^{-3-(-15)}=r^{-3 + 15}=r^{12}$. Wait, no, wait, original numerator exponent is -3, denominator exponent after applying power rule is -15. So $\frac{r^{-3}}{r^{-15}}=r^{-3-(-15)}=r^{12}$? Wait, no, the original problem's denominator is $2(r^{-3})^5$, numerator is $-2r^{-3}$. Wait, maybe I made a mistake earlier. Let's redo:

Wait, the expression is $\frac{-2r^{-3}}{2(r^{-3})^5}$. First, simplify the coefficients: $\frac{-2}{2}=-1$. Then for the exponents: $(r^{-3})^5 = r^{-15}$ (by power of a power: multiply exponents). Then we have $\frac{r^{-3}}{r^{-15}}$. Using the quotient rule: $a^m\div a^n=a^{m - n}$, so $r^{-3-(-15)}=r^{-3 + 15}=r^{12}$. Wait, but the previous attempt had $-1r^{-18}$, which is wrong. Wait, maybe I misread the exponent in the denominator. Wait, the denominator is $(r^{-3})^5$? Wait, no, maybe it's $(r^{3})^5$? No, the problem says $(r^{-3})^5$. Wait, let's check again.

Wait, the original problem: numerator is $-2r^{-3}$, denominator is $2(r^{-3})^5$. So coefficient: -2/2 = -1. Exponent part: $r^{-3}$ divided by $r^{-3\times5}=r^{-15}$. So $r^{-3}\div r^{-15}=r^{-3 - (-15)}=r^{12}$. So the correct simplification should be $-1\times r^{12}=-r^{12}$. Wait, but the user's attempt was $-1r^{-18}$, which is incorrect. So let's do it step by step correctly.

Step 1: Simplify coefficients: $\frac{-2}{2} = -1$.

Step 2: Simplify the power of $r$ in the denominator: $(r^{-3})^5 = r^{-3 \times 5} = r^{-15}$ (using $(a^m)^n = a^{mn}$).

Step 3: Simplify the fraction of the $r$ terms: $\frac{r^{-3}}{r^{-15}} = r^{-3 - (-15)}$ (using $\frac{a^m}{a^n} = a^{m - n}$) $= r^{-3 + 15} = r^{12}$.

So combining the coefficient and the variable part, we get $-1 \times r^{12} = -r^{12}$. Wait, but the user's attempt was $-1r^{-18}$, which is wrong. So maybe the denominator was $(r^{3})^5$? If the denominator was $(r^{3})^5$, then $(r^{3})^5 = r^{15}$, then $\frac{r^{-3}}{r^{15}}=r^{-3-15}=r^{-18}$, and coefficient -1, so $-r^{-18}$, which matches the attempt. Oh! Maybe there was a typo in the exponent sign in the denominator. If the denominator is $(r^{3})^5$ instead of $(r^{-3})^5$, then:

Denominator: $(r^{3})^5 = r^{15}$ (power of a power: 3*5=15). Then numerator exponent is -3, so $\frac{r^{-3}}{r^{15}}=r^{-3 - 15}=r^{-18}$. Then coefficient -1, so $-r^{-18}$. Ah, that must be the case. Maybe the original problem had a typo, or the user miswrote the exponent sign. So if the denominator is $(r^{3})^5$ (maybe a sign error), then the calculation would be:

Coefficient: -2/2 = -1. Exponent: $(r^{3})^5 = r^{15}$, so $\frac{r^{-3}}{r^{15}}=r^{-3-15}=r^{-18}$. Then the result is $-1\times r^{-18}=-r^{-18}$, which is what the attempt had. But if the denominator is $(r^{-3})^5$, then the correct exponent is 12. So probably there was a sign error in the denominator's exponent. Assuming the denominator is $(r^{3})^5$ (maybe a typo, and the exponent is positive 3), then the steps are:

Step 1: Simplify coefficients: $\frac{-2}{2}=-1$.

Step 2: Simplify the exponent in the denominator: $(r^{3})^5 = r^{3\times5}=r^{15}$ (power of a power rule).

Step 3: Simplify the variable fraction: $\frac{r^{-3}}{r^{15}} = r^{-3 - 15}=r^{-18}$ (quotient rule: subtract exponents).

So combining, we get $-1\times r^{-18}=-r^{-18}$. But wait, the problem says "only using positive exponents". Oh! Right, we need to convert negative exponents to positive by using $a^{-n}=\frac{1}{a^n}$. So $r^{-18}=\frac{1}{r^{18}}$, so the expression becomes $-\frac{1}{r^{18}}$. Wait, but the problem says "only using positive exponents", so we need to rewrite $r^{-18}$ as $\frac{1}{r^{18}}$. So the correct simplification (if the denominator exponent was positive 3, maybe a typo) would be:

First, coefficients: -2/2 = -1.

Exponent part: $(r^{3})^5 = r^{15}$ (if it was positive 3, but the problem says $(r^{-3})^5$). Wait, this is confusing. Let's start over with the given problem:

Given: $\frac{-2r^{-3}}{2(r^{-3})^5}$

Step 1: Simplify coefficients: $\frac{-2}{2} = -1$.

Step 2: Simplify the power in the denominator: $(r^{-3})^5 = r^{-3 \times 5} = r^{-15}$ (power of a power: multiply exponents).

Step 3: Simplify the variable fraction: $\frac{r^{-3}}{r^{-15}}$ using quotient rule: $a^m \div a^n = a^{m - n}$, so $r^{-3 - (-15)} = r^{-3 + 15} = r^{12}$.

Step 4: Combine coefficient and variable: $-1 \times r^{12} = -r^{12}$.

But the user's attempt was $-1r^{-18}$, which is wrong. So where is the mistake? Ah! Wait, maybe the denominator is $(r^{3})^5$ instead of $(r^{-3})^5$? If that's the case:

Denominator: $(r^{3})^5 = r^{15}$.

Then variable fraction: $\frac{r^{-3}}{r^{15}} = r^{-3 - 15} = r^{-18}$.

Then coefficient: -1.

So the expression is $-1r^{-18}$, but we need positive exponents, so $-1r^{-18} = -\frac{1}{r^{18}}$.

But the problem says "only using positive exponents", so we must rewrite negative exponents as positive. So the correct form with positive exponents is $-\frac{1}{r^{18}}$. But the user's attempt has $-1r^{-18}$, which still has a negative exponent, so it's incorrect.

Wait, maybe the original problem's denominator was $(r^{3})^5$ (a typo, missing the negative sign), which would make the exponent in the denominator positive 15, leading to $r^{-3 - 15}=r^{-18}$, then converting to positive exponent gives $-\frac{1}{r^{18}}$.

Alternatively, if the problem is correct as written (denominator is $(r^{-3})^5$), then:

After step 3, we have $r^{12}$, so the expression is $-r^{12}$, which has a positive exponent, so that's correct.

So there are two possibilities depending on whether the denominator's exponent was a typo. But given the user's attempt was $-1r^{-18}$, they probably used $(r^{3})^5$ (positive 3) instead of $(r^{-3})^5$. So to correct the simplification with positive exponents:

If we have $-1r^{-18}$, to convert to positive exponents, we use $a^{-n}=\frac{1}{a^n}$, so $-1r^{-18}=-\frac{1}{r^{18}}$.

But let's follow the problem's instructions: "only using positive exponents". So we need to rewrite any negative exponents as positive.

So let's do it correctly with the given problem (denominator is $(r^{-3})^5$):

1. Coefficients: $\frac{-2}{2}=-1$.

2. Exponent in denominator: $(r^{-3})^5 = r^{-15}$ (power of a power: multiply exponents).

3. Variable fraction: $\frac{r^{-3}}{r^{-15}} = r^{-3 - (-15)} = r^{12}$ (quotient rule: subtract exponents).

4. Combine: $-1 \times r^{12} = -r^{12}$. This has a positive exponent, so it's in simplest form with positive exponents.

But the user's attempt was $-1r^{-18}$, which is wrong. So the mistake was in the exponent calculation for the denominator. They probably thought the exponent in the denominator was $(r^{3})^5$ (positive 3), leading to $r^{15}$, then $r^{-3 - 15}=r^{-18}$. Then, to convert to positive exponents, it's $-\frac{1}{r^{18}}$.

So depending on the problem's actual exponent (maybe a typo), but given the user's attempt, we can see where they went wrong. But the correct simplification (if the denominator is $(r^{-3})^5$) is $-r^{12}$, with positive exponent. If the denominator was $(r^{3})^5$, then it's $-\frac{1}{r^{18}}$.

But let's proceed with the given problem as written: $\frac{-2r^{-3}}{2(r^{-3})^5}$

Simplify step by step:

- Coefficients: $\frac{-2}{2} = -1$.

- Denominator exponent: $(r^{-3})^5 = r^{-3 \times 5} = r^{-15}$.

- Variable part: $\frac{r^{-3}}{r^{-15}} = r^{-3 - (-15)} = r^{12}$ (because $-3 - (-15) = 12$).

- So the expression becomes $-1 \times r^{12} = -r^{12}$.

This is in simplest form with a positive exponent, so that's the answer. The user's attempt had a mistake in the exponent calculation (probably used positive 3 instead of negative 3 in the denominator's exponent), leading to $r^{-18}$ instead of $r^{12}$.

Answer:

$-r^{12}$ (or $-\dfrac{1}{r^{-12}}$ but no, $r^{12}$ is positive exponent, so $-r^{12}$ is correct with positive exponent. If we must write without the coefficient's 1, it's $-r^{12}$).