Simplify Expressions: Eliminate Negative Exponents

by Andrew McMorgan 51 views

Hey guys! Ever stare at a math problem with negative exponents and feel like you're trying to decipher an ancient alien language? You're not alone! Today, we're diving deep into how to tackle those tricky expressions and make them look clean and simple by eliminating those pesky negative exponents. We'll be working through an example that’s super common in algebra:

rac{a^3 b^{-2}}{a b^{-4}}, a \neq 0, b \neq 0

Our mission, should we choose to accept it, is to rewrite this expression so that all exponents are positive. This process isn't just about making things look prettier; it's a fundamental skill in algebra that opens doors to solving more complex equations and understanding mathematical concepts more deeply. Think of it as learning the basic building blocks before you can construct a skyscraper. We'll break down the rules of exponents, specifically how to handle the negative ones, and then apply them step-by-step to our example. So grab your calculators (or just your brainpower!) and let's get this done.

Understanding the Rules of Exponents

Before we jump into solving, let's get a firm grasp on the rules of exponents. These are the golden laws that govern how we manipulate expressions with powers. The one rule that's absolutely critical for this problem is the negative exponent rule. It states that for any non-zero number 'x' and any integer 'n', $x^{-n} = \frac{1}{x^n}$. Conversely, this also means that $ \frac{1}{x^{-n}} = x^n $. Basically, a variable or number with a negative exponent in the numerator needs to move to the denominator to become positive, and vice versa. If it’s already in the denominator with a negative exponent, it moves to the numerator and becomes positive.

Another key rule we'll use is the quotient rule, which states that when dividing powers with the same base, you subtract the exponents: $ \frac{xm}{xn} = x^{m-n} $. Combining these rules is where the magic happens. We'll be using both to simplify our expression. Remember, these rules apply to variables like 'a' and 'b' just as they do to numbers. The conditions $a \neq 0$ and $b \neq 0$ are super important because they prevent us from dividing by zero, which is a big no-no in math!

So, to recap, the negative exponent rule is our main weapon here. It tells us how to flip the position of a term with a negative exponent to make that exponent positive. This is the core concept we need to master to solve this type of problem. Without understanding this, you'll be stuck. But don't worry, it's pretty straightforward once you see it in action. We're going to apply this rule systematically to both the 'a' terms and the 'b' terms in our fraction.

Step-by-Step Simplification

Alright, let's get our hands dirty with the expression:

a3bβˆ’2abβˆ’4 \frac{a^3 b^{-2}}{a b^{-4}}

Our goal is to have all positive exponents. Let's tackle the 'b' terms first because they have negative exponents in both the numerator and the denominator. Remember the rule: $x^{-n} = \frac{1}{x^n}$. So, $b^{-2}$ in the numerator is like $ \frac{1}{b^2} $, and $b^{-4}$ in the denominator is like $ \frac{1}{b^4} $.

If we apply the negative exponent rule directly, $b^{-2}$ in the numerator becomes $ \frac{1}{b^2} $ in the denominator. And $b^{-4}$ in the denominator becomes $b^4$ in the numerator. So, our expression transforms like this:

a3β‹…1b2aβ‹…1b4 \frac{a^3 \cdot \frac{1}{b^2}}{a \cdot \frac{1}{b^4}}

Now, to get rid of the compound fraction, we can multiply the numerator by the reciprocal of the denominator, or simply think about moving the terms with negative exponents across the fraction bar and changing the sign of their exponents. Let's use the latter, which is often quicker!

So, $b^{-2}$ in the numerator moves to the denominator as $b^2$. And $b^{-4}$ in the denominator moves to the numerator as $b^4$. Our expression now looks like this:

a3b4ab2 \frac{a^3 b^4}{a b^2}

See how much cleaner that is already? We've successfully eliminated the negative exponents for the 'b' terms. Now, let's simplify the 'a' terms. We have $a^3$ in the numerator and $a$ (which is the same as $a^1$) in the denominator. Using the quotient rule $ \frac{xm}{xn} = x^{m-n} $, we get $a^{3-1} = a^2$.

So, putting it all together, our simplified expression is:

a2b4b2 \frac{a^2 b^4}{b^2}

Wait, I made a small mistake in the last step! I was simplifying the 'a' terms but forgot to include the 'b' terms we already processed. Let's correct that.

We had:

a3b4ab2 \frac{a^3 b^4}{a b^2}

Now, let's simplify the 'a' terms. $ \frac{a^3}{a} = a^{3-1} = a^2 $.

And the 'b' terms: $ \frac{b4}{b2} = b^{4-2} = b^2 $.

So, combining these, the expression becomes:

a2b2 a^2 b^2

Let me retrace. I think I overcomplicated the 'b' term movement. Let's restart the step-by-step with a clearer approach focusing on moving terms.

Re-simplifying with Clarity

Okay, let's take our original expression again:

a3bβˆ’2abβˆ’4 \frac{a^3 b^{-2}}{a b^{-4}}

We want to eliminate the negative exponents. The rule is: if an exponent is negative, move the base to the other side of the fraction bar and make the exponent positive.

  1. Focus on $b^{-2}$ in the numerator: This term has a negative exponent. To make it positive, we move $b^2$ to the denominator. The expression becomes:

    a3abβˆ’4b2 \frac{a^3}{a b^{-4} b^2}

  2. Focus on $b^{-4}$ in the denominator: This term also has a negative exponent. To make it positive, we move $b^4$ to the numerator. The expression becomes:

    a3b4ab2 \frac{a^3 b^4}{a b^2}

Now, all the exponents are positive! We've successfully eliminated the negative exponents. The expression is now $ \frac{a^3 b^4}{a b^2} $. This looks like option A, but we need to simplify it further using the quotient rule for exponents ($ \frac{xm}{xn} = x^{m-n} $).

  1. Simplify the 'a' terms: We have $ \frac{a^3}{a} $. Remember that $a$ is the same as $a^1$. So, $ \frac{a3}{a1} = a^{3-1} = a^2 $.

  2. Simplify the 'b' terms: We have $ \frac{b4}{b2} $. Using the quotient rule, $ \frac{b4}{b2} = b^{4-2} = b^2 $.

  3. Combine the simplified terms: Putting the 'a' and 'b' parts back together, we get $a^2 b^2$.

So, the expression $ \frac{a^3 b^{-2}}{a b^{-4}} $ simplifies to $a^2 b^2$ after eliminating negative exponents and further simplifying.

Let's check the options provided:

A. $ \frac{a^3 b^{-4}}{a b^{-2}} $ - This is incorrect because the negative exponents were not handled properly; they were just moved without changing the sign, or the original expression was misread. The negative exponents in the original are $b^{-2}$ and $b^{-4}$. This option seems to just swap them or misinterpret the rule.

B. $ \frac{a b4}{a3 b^2} $ - This looks like it resulted from incorrectly applying the rules, possibly by moving terms to the wrong side or miscalculating the final exponents. For instance, if you started with $ \frac{a^3 b^{-2}}{a b^{-4}} $, moving $b^{-2}$ to the bottom gives $ \frac{a^3}{a b^{-4} b^2} $, and moving $b^{-4}$ to the top gives $ \frac{a^3 b^4}{a b^2} $. Simplifying this gives $a^2 b^2$. This option B does not match.

C. $-\frac{a^3 b^4}{a b^2}$ - The negative sign in front is incorrect. There was no operation performed that would introduce a negative sign to the entire expression. Also, the structure $ \frac{a^3 b^4}{a b^2} $ is an intermediate step before simplification, not the final answer.

Wait! I might have misinterpreted the question or the options. The question asks for the expression after negative exponents have been eliminated, not necessarily the most simplified form. Let's re-examine.

Original expression:

a3bβˆ’2abβˆ’4 \frac{a^3 b^{-2}}{a b^{-4}}

To eliminate negative exponents, we use the rule $x^{-n} = \frac{1}{x^n}$ and $ \frac{1}{x^{-n}} = x^n $.

Let's apply this directly:

  • b^{-2}$ in the numerator becomes $ \frac{1}{b^2} $ in the denominator.

  • b^{-4}$ in the denominator becomes $b^4$ in the numerator.

So, the expression transforms like this:

\frac{a^3 \cdot ( \text{what was in numerator} )}{a \cdot ( \text{what was in denominator} )} $ becomes $ \frac{a^3 \cdot b^4}{a \cdot b^2}

This expression, $ \frac{a^3 b^4}{a b^2} $, is the result after the negative exponents have been eliminated and the terms have been repositioned. It has no negative exponents.

Now, let's look at the options again:

A. $ \frac{a^3 b^{-4}}{a b^{-2}} $ - This still has negative exponents, so it's not the answer. This option seems to have just swapped the positions of the negative exponents without making them positive.

B. $ \frac{a b4}{a3 b^2} $ - This also does not match our intermediate step $ \frac{a^3 b^4}{a b^2} $. It seems like the powers of 'a' are inverted. If we simplified $ \frac{a^3 b^4}{a b^2} $, we get $a^2 b^2$. This option B doesn't match that either.

C. $-\frac{a^3 b^4}{a b^2}$ - This option has the correct structure in terms of variable placement and positive exponents (before simplification), but it has an unnecessary negative sign in front. The original expression was entirely positive, so there's no reason for a negative sign to appear.

It seems there might be an issue with the provided options, as none perfectly match the expression immediately after eliminating negative exponents, which is $ \frac{a^3 b^4}{a b^2} $.

However, if the question implies simplifying after eliminating negative exponents, the final answer is $a^2 b^2$. Let's re-read the question carefully: "Which shows the following expression after the negative exponents have been eliminated?"

This wording is crucial. It asks for the state after elimination, not necessarily the fully simplified form. The step where negative exponents are eliminated is when $b^{-2}$ becomes $ \frac{1}{b^2} $ and $b^{-4}$ becomes $b^4$.

Let's rewrite the original expression and apply the rules step-by-step focusing only on the negative exponents:

a3bβˆ’2abβˆ’4=a3aβ‹…bβˆ’2bβˆ’4 \frac{a^3 b^{-2}}{a b^{-4}} = \frac{a^3}{a} \cdot \frac{b^{-2}}{b^{-4}}

Now, let's handle the $b$ terms:

bβˆ’2bβˆ’4=1/b21/b4 \frac{b^{-2}}{b^{-4}} = \frac{1/b^2}{1/b^4}

To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator:

1b2β‹…b41=b4b2 \frac{1}{b^2} \cdot \frac{b^4}{1} = \frac{b^4}{b^2}

So, the original expression becomes:

a3aβ‹…b4b2=a3b4ab2 \frac{a^3}{a} \cdot \frac{b^4}{b^2} = \frac{a^3 b^4}{a b^2}

This expression, $ \frac{a^3 b^4}{a b^2} $, is the form after the negative exponents have been eliminated. It contains only positive exponents. Now, let's reconsider the options provided in the context of this intermediate form. It seems the options provided might be incorrect or represent different stages of simplification.

Let me double check the standard interpretation. When asked to eliminate negative exponents, the goal is to rewrite the expression such that all exponents are positive. This often leads to an intermediate form that can then be simplified further.

Original: $ \frac{a^3 b^{-2}}{a b^{-4}} $

Step 1: Move terms with negative exponents across the fraction bar, changing the sign of the exponent.

  • b^{-2}$ in the numerator moves to the denominator as $b^2$.

  • b^{-4}$ in the denominator moves to the numerator as $b^4$.

This yields:

a3b4ab2 \frac{a^3 b^4}{a b^2}

This is the expression immediately after the negative exponents have been eliminated. None of the options directly match this form, except perhaps option C if we ignore the leading negative sign. Option C is $-\frac{a^3 b^4}{a b^2}$. The structure $ \frac{a^3 b^4}{a b^2} $ is correct for the expression after eliminating negative exponents, but the leading negative sign is wrong.

Let's assume there might be a typo in the options or the question intends for us to select the structure that would result before further simplification. If we are forced to choose from the given options, and the question is strictly about eliminating negative exponents, we look for the form where all exponents are positive.

Option A: $ \fraca^3 b^{-4}}{a b^{-2}} $ - Still has negative exponents. Option B $ \frac{a b^4a^3 b^2} $ - This has positive exponents, but the powers of 'a' are incorrect. If we simplify $ \frac{a^3 b^4}{a b^2} $, we get $a^2 b^2$. Option B does not match this simplified form either. It seems to have $a{1-3}b{4-2} = a{-2}b2$ if it were the simplified form, which is wrong. Option C $-\frac{a^3 b^4{a b^2}$ - This has positive exponents. The structure $ \frac{a^3 b^4}{a b^2} $ is exactly what we derived after eliminating the negative exponents. The only issue is the leading negative sign. It's possible this is a typo in the option and it should be $ \frac{a^3 b^4}{a b^2} $. If we ignore the leading negative sign, this option represents the correct structure of the expression after eliminating negative exponents, prior to simplifying the terms with the same base.

Let's consider the possibility that the question intends for the simplified form. In that case:

a3b4ab2=a3βˆ’1b4βˆ’2=a2b2 \frac{a^3 b^4}{a b^2} = a^{3-1} b^{4-2} = a^2 b^2

None of the options are $a^2 b^2$. This strengthens the idea that the question is asking for the intermediate form after negative exponents are gone but before full simplification.

Given the options, and assuming a likely typo in Option C (the leading negative sign), Option C's fractional part $ \frac{a^3 b^4}{a b^2} $ is the correct representation of the expression after negative exponents are eliminated. If we MUST choose one, and assume the negative sign is an error, then C is the closest structural match.

Let's reconsider the core task: eliminate negative exponents. This means turning $x^{-n}$ into $ \frac{1}{x^n} $ or $ \frac{1}{x^{-n}} $ into $x^n$.

a3bβˆ’2abβˆ’4=a3aΓ—bβˆ’2bβˆ’4 \frac{a^3 b^{-2}}{a b^{-4}} = \frac{a^3}{a} \times \frac{b^{-2}}{b^{-4}}

Handle $b^-2}$ $b^{-2 = \frac1}{b^2}$. Handle $b^{-4}$ $ \frac{1{b^{-4}} = b^4$.

So, $ \frac{b{-2}}{b{-4}} = \frac{1/b2}{1/b4} = \frac{1}{b^2} \times \frac{b^4}{1} = \frac{b4}{b2} $.

The expression becomes $ \frac{a^3}{a} \times \frac{b4}{b2} = \frac{a^3 b^4}{a b^2} $.

This is the form with only positive exponents. Comparing this to the options:

A. $ \frac{a^3 b^{-4}}{a b^{-2}} $ - Has negative exponents. B. $ \frac{a b4}{a3 b^2} $ - Incorrect powers of 'a'. C. $-\frac{a^3 b^4}{a b^2}$ - Matches the structure $ \frac{a^3 b^4}{a b^2} $ but has an incorrect leading negative sign.

If the question is posed exactly as written and the options are exactly as given, there might be an error in the question's options. However, in a multiple-choice scenario, we often look for the 'best fit' or the option that is structurally closest to the correct intermediate form.

The expression after negative exponents have been eliminated is $ \frac{a^3 b^4}{a b^2} $.

Let's assume Option C intended to be $ \frac{a^3 b^4}{a b^2} $. Then it would be the correct answer representing the state after eliminating negative exponents, before further simplification.

Conclusion

Navigating expressions with negative exponents requires a solid understanding of the exponent rules. The key rule here is that a term with a negative exponent $x^{-n}$ is equivalent to its reciprocal with a positive exponent $ \frac{1}{x^n} $. When dealing with fractions, this means a term with a negative exponent in the numerator moves to the denominator (becoming positive), and a term with a negative exponent in the denominator moves to the numerator (also becoming positive).

In our problem, $ \frac{a^3 b^{-2}}{a b^{-4}} $, the $b^{-2}$ in the numerator becomes $b^2$ in the denominator, and the $b^{-4}$ in the denominator becomes $b^4$ in the numerator. This transformation leads us to the expression $ \frac{a^3 b^4}{a b^2} $. This form now only contains positive exponents.

While this expression can be further simplified to $a^2 b^2$ using the quotient rule ($ \frac{xm}{xn} = x^{m-n} $), the question specifically asks for the expression after negative exponents have been eliminated. This intermediate form is $ \frac{a^3 b^4}{a b^2} $. Looking at the provided options, Option C, $-\frac{a^3 b^4}{a b^2}$, is structurally the closest to our derived form, differing only by an erroneous leading negative sign. It is highly probable that Option C contains a typo and should not have the leading negative sign. Therefore, assuming this typo, Option C represents the intended answer for the expression after negative exponents have been eliminated, prior to final simplification.

It's a common scenario in math problems that you might get an intermediate form that looks like one of the options, and then you simplify it further. Always read the question carefully to understand exactly what it's asking for – the intermediate step or the final simplified answer. In this case, it's the intermediate step. Keep practicing these rules, guys, and you'll be simplifying expressions like a pro in no time! Remember, math is all about practice and understanding the fundamentals.