Multiplication of Powers: Number of terms

$2^{10}\cdot2^7\cdot2^6=$

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.

$2^{23}$

$8^2\cdot8^3\cdot8^5=$

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

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

$8^{10}$

$(y^3\times x^2)^4=$

We will solve the problem in two steps, in the first step we will use the power of a product rule:

$(z\cdot t)^n=z^n\cdot t^n$The rule states that the power affecting a product within parentheses applies to each of the elements of the product when the parentheses are opened,

We begin by applying the law to the given problem:

$(y^3\cdot x^2)^4=(y^3)^4\cdot(x^2)^4$When we open the parentheses, we apply the power to each of the terms of the product separately, but since each of these terms is already raised to a power, we must be careful to use parentheses.

We then use the power of a power rule.

$(b^m)^n=b^{m\cdot n}$We apply the rule to the given problem and we should obtain the following result:

$(y^3)^4\cdot(x^2)^4=y^{3\cdot4}\cdot x^{2\cdot4}=y^{12}\cdot x^8$When in the second step we perform the multiplication operation on the power exponents of the obtained terms.

Therefore, the correct answer is option d.

$y^{12}x^8$

Simplify the following:

$\lbrack\frac{a^4}{a^3}\times\frac{a^8}{a^7}\rbrack:\frac{a^{10}}{a^8}$

$1$

Simplify the following:

$\frac{a^{12}}{a^9}\times\frac{a^3}{a^4}=$

$a^2$

$((x^{\frac{1}{4}}\times3^2\times6^3)^{\frac{1}{4}})^8=$

$x^{\frac{1}{2}}\times3^4\times6^6$

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=\text{?}$

$\frac{1}{9^{-3}}$

$\frac{1}{-3}\cdot3^{-4}\cdot5^3=\text{?}$

$-\frac{5^3}{3^5}$

$45^{-80}\cdot\frac{1}{45^{-81}}\cdot49\cdot7^{-5}=\text{?}$

$\frac{45}{7^3}$

Solve the following:

$\frac{y^3}{y^6}\times\frac{y^4}{y^{-2}}\times\frac{y^{12}}{y^7}=$

$y^8$

Factor the following expression:

$2a^5+8a^6+4a^3$

$a^3(25a^2+8a^3+4)$

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

$a^8\times b^8$

$b^{-3}\times b^3\times b^4\times b^{-2}=$

$b^2$

$7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=\text{?}$

$\frac{4^{-1}3^{-7}}{7^{-2}}$

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

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

Simplify the following expression:

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

$378^3+9^1+7^{60}+2^4$