Multiplication of Powers: Number of terms

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Let's return to the problem:

Keep in mind that all the terms in the multiplication have the same base, so we will use the previous property:

$2^{10}\cdot2^7\cdot2^6=2^{10+7+6}=2^{23}$Therefore, the correct answer is option c.

All bases are equal and therefore the exponents can be added together.

$8^2\cdot8^3\cdot8^5=8^{10}$

Let's solve this in two stages. In the first stage, we'll use the power rule for powers in parentheses:

$(z\cdot t)^n=z^n\cdot t^n$

which states that when a power is applied to terms in parentheses, it applies to each term inside the parentheses when they are opened,

Let's apply this rule to our problem:

$\big((x^{\frac{1}{4}}\cdot3^2\cdot6^3)^{\frac{1}{4}}\big)^8=((x^{\frac{1}{4}})^{\frac{1}{4}}\cdot(3^2)^{\frac{1}{4}}\cdot(6^3)^{\frac{1}{4}})^8$

where when opening the parentheses, we applied the power to each term separately, but since each of these terms is raised to a power, we did this carefully and used parentheses,

Next, we'll use the power rule for a power raised to a power:

$(b^m)^n=b^{m\cdot n}$

Let's apply this rule to the expression we got:

$(x^{\frac{1}{4}\cdot\frac{1}{4}}\cdot3^{2\cdot\frac{1}{4}}\cdot6^{3\cdot\frac{1}{4}})^8=(x^{\frac{1}{16}}\cdot3^{\frac{2}{4}}\cdot6^{\frac{3}{4}})^8=x^{\frac{1}{16}\cdot8}\cdot3^{\frac{2}{4}\cdot8}\cdot6^{\frac{3}{4}\cdot8}=x^{\frac{8}{16}}\cdot3^{\frac{16}{4}}\cdot6^{\frac{24}{4}}$

where in the second stage we performed multiplication in the fractions of the power expressions of the terms we obtained, remembering that multiplication in fractions is actually multiplication in the numerator, and then - in the final stage we simplified the fractions in the power expressions of the multiplication terms we got:

$x^{\frac{8}{16}}\cdot3^{\frac{16}{4}}\cdot6^{\frac{24}{4}}=x^{\frac{1}{2}}\cdot3^4\cdot6^6$

Therefore, the correct answer is answer B.

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.

In solving the problem we will use two laws of exponents, let's recall them:

a. The law of exponents for multiplying terms with identical bases:

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

b. The law of exponents for power of a power:

$(a^m)^n=a^{m\cdot n}$

We will apply these two laws of exponents to the expression in the problem in two stages:

We'll start by applying the law of exponents mentioned in a to the second term from the left in the expression:

$9^{-3}\cdot9^4=9^{-3+4}=9^1=9$

After we applied the law of exponents mentioned in a in the first stage and simplified the resulting expression,

We'll continue to the next stage and apply the law of exponents mentioned in b and deal with the third term from the left in the expression, we'll do this in two steps:

$((7^2)^5)^6=(7^2)^{5\cdot6}=7^{2\cdot5\cdot6}=7^{60}$

When in the first step we applied the law of exponents mentioned in b and eliminated the outer parentheses, in the next step we applied the same law of exponents again and eliminated the remaining parentheses, in the following steps we simplified the resulting expression,

Let's summarize the two stages detailed above for the complete solution of the problem:

$(9\cdot7\cdot6)^3+9^{-3}\cdot9^4+((7^2)^5)^6+2^4 = (9\cdot7\cdot6)^3+9+7^{60}+2^4$

In the next step we'll calculate the result of multiplying the terms inside the parentheses in the leftmost term:

$(9\cdot7\cdot6)^3+9+7^{60}+2^4 =378^3+9+7^{60}+2^4$

Therefore the correct answer is answer d.

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.

First, we'll use the distributive property of multiplication and arrange the algebraic expression according to like bases:

$a^{10}\cdot a^{-2}\cdot b^5\cdot b^3$

From here on we will no longer indicate the multiplication sign, but instead use the conventional writing method where placing terms next to each other means multiplication.

Note that we need to multiply terms with identical bases, so we'll use the appropriate power rule:

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

Note that this rule can only be used to calculate multiplication between terms with identical bases,

In this problem, there is also a term with a negative exponent, but this doesn't pose an issue regarding the use of the aforementioned power rule. In fact, this power rule is valid in all cases for numerical terms with different exponents, including negative exponents, rational exponents, and even irrational exponents, etc.,

Let's apply it to the problem:

$a^{10}a^{-2}b^5b^3=a^{10-2}b^{5+3}=a^8b^8$

When we dealt separately with terms having equal bases, meaning separately with terms having bases $a$ and $b$

therefore the correct answer is D.

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

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

Note that this law is valid for any number of terms in multiplication and not just for two terms. For example, for multiplication of three terms with the same base we get:

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

When we used the above exponent law twice, we can also perform the same calculation for four terms in multiplication five and so on...

Additionally, note that this law can only be used to calculate multiplication performed between terms with identical bases,

In this problem there are also terms with negative exponents, but this doesn't pose an issue regarding the use of the above 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.

From here on we will no longer indicate the multiplication sign, but use the conventional writing form where placing terms next to each other means multiplication.

Let's return to the problem and apply the above law:

$b^{-3}b^3b^4b^{-2}=b^{-3+3+4-2}=b^2$

Therefore the correct answer is B.