Multiplication of Powers: Calculating powers with negative exponents

We begin by using the rule for multiplying exponents. (the multiplication between terms with identical bases):

$a^m\cdot a^n=a^{m+n}$We then apply it to the problem:

$7^5\cdot7^{-6}=7^{5+(-6)}=7^{5-6}=7^{-1}$When in a first stage we begin by applying the aforementioned rule and then continue on to simplify the expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression obtained in the previous step:

$7^{-1}=\frac{1}{7^1}=\frac{1}{7}$We then summarise the solution to the problem: $7^5\cdot7^{-6}=7^{-1}=\frac{1}{7}$Therefore, the correct answer is option B.

Note that we need to calculate multiplication between terms with identical bases, so we'll use the appropriate exponent law:

$b^m\cdot b^n=b^{m+n}$Note that we can only use this law to calculate multiplication performed between terms with identical bases,

Here in the problem there is also a term with a negative exponent, but this does not pose an issue regarding the use of the aforementioned exponent law. In fact, this exponent law is valid in all cases for numerical terms with different exponents, including negative exponents, rational number exponents, and even irrational number exponents, etc.,

Let's apply it to the problem:

$y^{-2}\cdot y^7=y^{-2+7}=y^5$Therefore the correct answer is A.

We begin by using the power rule of exponents; for the multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it to the given problem:

$12^4\cdot12^{-6}=12^{4+(-6)}=12^{4-6}=12^{-2}$When in a first stage we apply the aforementioned rule and then simplify the subsequent expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression that we obtained in the previous step:

$12^{-2}=\frac{1}{12^2}=\frac{1}{144}$Lastly we summarise the solution to the problem: $12^4\cdot12^{-6}=12^{-2} =\frac{1}{144}$Therefore, the correct answer is option A.

First we'll use the laws of exponents for negative exponents, but in the opposite direction:

$\frac{1}{a^n} = a^{-n}$

and we'll handle using the middle term in the multiplication in the problem:

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=9^{300}\cdot9^{-(-252)}\cdot9^{-549}=9^{300}\cdot9^{252}\cdot9^{-549}$

where in the first stage we'll apply the aforementioned law of exponents, and this carefully since the term in the denominator of the fraction has a negative exponent, therefore we used parentheses, then we simplified the expression in the exponent,

Next we'll recall the law of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and we'll apply this law to the last expression we got:

$9^{300}\cdot9^{252}\cdot9^{-549}=9^{300+252+(-549)}=9^{300+252-549}=9^3$

Let's summarize the steps so far, we got:

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=9^{300}\cdot9^{252}\cdot9^{-549} =9^{300+252-549}=9^3$

Note that there isn't such an answer among the answer choices, however we can always represent the expression we got as a term with a negative exponent by taking the minus sign outside the parentheses in the exponent, meaning we'll do:

$9^3=9^{-(-3)}$

and then we'll use again the law of negative exponents:

$a^{-n}=\frac{1}{a^n}$

Let's apply it to the last expression we got:

$9^3=9^{-(-3)}=\frac{1}{9^{-3}}$

Therefore we got that:

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=9^3 =9^{-(-3)}=\frac{1}{9^{-3}}$

And therefore the correct answer is answer A.

First we'll use the laws of exponents for negative exponents, but in the opposite direction:

$\frac{1}{a^n} = a^{-n}$

and we'll handle using the middle term in the multiplication in the problem:

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2}=4^{2x}\cdot4^{-1}\cdot4^{-2}$

Next, we'll recall the law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and we'll apply this law to the last expression we got:

$4^{2x}\cdot4^{-1}\cdot4^{-2}=4^{2x+(-1)+(-2)}=4^{2x-1-2}=4^{2x-3}$

We got the most simplified expression,

Let's summarize the steps so far, we got:

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2}=4^{2x}\cdot4^{-1}\cdot4^{-2} =4^{2x-3}$

A quick look at the options will reveal that there isn't such an answer among the options and another check of what we've done so far will show that there are no calculation errors,

This means that another mathematical manipulation is needed on the expression we got, a hint for the required manipulation could be the fact that answer D is similar to our expression but the exponent has a minus sign compared to the exponent we got in the final expression and the expression itself is in a fraction where the numerator is 1, which reminds us of the negative exponent law, let's check this suspicion and handle the expression we got in the following way:

$4^{2x-3}=4^{-(-2x+3)}=4^{-(3+(-2x))}=4^{-(3-2x)}$

where the goal is to present the expression we got in the form of a term with a negative exponent, we did this by taking the minus sign outside the parentheses in the exponent and rearranging the expression inside the parentheses using the commutative law of addition and then simplified the expression in parentheses,

Now let's use the negative exponent law again:

$a^{-n} =\frac{1}{a^n}$

And apply it to the expression we got:

$4^{-(3-2x)}=\frac{1}{4^{3-2x}}$

We got therefore that the expression we got earlier can be written as:

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2} =4^{2x-3} = 4^{-(3-2x)}=\frac{1}{4^{3-2x}}$

Therefore the correct answer is indeed answer D.