Power of a Power: Number of terms

$(2^2)^3+(3^3)^4+(9^2)^6=$

We use the formula:

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

$(2^2)^3+(3^3)^4+(9^2)^6=2^{2\times3}+3^{3\times4}+9^{2\times6}=2^6+3^{12}+9^{12}$

$2^6+3^{12}+9^{12}$

Solve the exercise:

$(x^2\times3)^2=$

We have an exponent raised to another exponent with a multiplication between parentheses:

$(z\cdot t)^n=z^n\cdot t^n$This says that in a case where a power is applied to a multiplication between parentheses,the power is applied to each term of the multiplication when the parentheses are opened,

We apply it in the problem:

$(3x^2)^2=3^2(x^2)^2$With the second term of the multiplication we proceed carefully, since it is already in a power (that's why we use parentheses). The term will be raised using the power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$and we apply it in the problem:

$3^2(x^2)^2=9x^{2\cdot2}=9x^4$In the first step we raise the number to the power, and in the second step we multiply the exponent.

Therefore, the correct answer is option a.

$9x^4$

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

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.

$y^{12}x^8$

$((8by)^3)^y+(3^x)^a=$

$\left(8by\right)^{3\cdot y}+3^{x\cdot a}$

We begin by applying the following rule:

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

We then open the parentheses according to the above rule.

$\left(abc\right)^x=a^x\cdot b^x\cdot c^x$

$8^{3y}\cdot b^{3y}\cdot y^{3y}+3^{xa}$

$8^{3y}\times b^{3y}\times y^{3y}+3^{ax}$

$((7\times3)^2)^6+(3^{-1})^3\times(2^3)^4=$

$21^{12}+3^{-3}\times2^{12}$

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

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

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

$x^4y^6z^8$

$((9xyz)^3)^4+(a^y)^x=$

$9^{12}x^{12}y^{12}z^{12}+a^{yx}$

Simplify the following expression:

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

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