When we have an expression raised to a power that, in turn, is raised (within parentheses) to another power, we can multiply the exponents and raise the base number to the result of this multiplication.

When we have an expression raised to a power that, in turn, is raised (within parentheses) to another power, we can multiply the exponents and raise the base number to the result of this multiplication.

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

This property is also concerning algebraic expressions.

\( [(4-2)^2]^3= \)

$(4^3 )^2=$

We can see that the exponent $2$ applies to the entire expression $4^3$.

therefore, we can multiply both exponents and raise the base to the result of the multiplication.** We will obtain:**

$4^{3\times2}=4^6=4096$

If we were presented with an exercise in which there is a certain power over a term that already has another power, we will multiply the powers that have equal bases.

**Let's start with an easy one:**

**$(X^{6-3})^4=$**

We'll see that there is a subtraction in the exponents of $X$ and that, first, we must deal with it.

**We'll do this and obtain:**

$(X^3)^4=$

Now we can apply the power of a power property and multiply the exponents, we will obtain:

$X^{12}$

Good. **Let's move on to a more complicated example:**

**$(\frac{2X^2}{X})^4\cdot(\frac{4Y^2}{Y})^3=$**

**Recommendation:**

Before applying the power located outside the parentheses to each of the terms separately, first, it is advisable to carefully observe the exercise.

Upon observing it, you will realize that you can reduce or subtract exponents from the fractions themselves, before touching the exponent located outside the parentheses.

We will subtract the exponents of the corresponding bases (we will reduce) and obtain:

$(2X)^4\cdot(4Y)^3=$

Now we can apply the exponent to each of the terms separately (do not forget about the coefficients) and we will get:

$16X^4\cdot64Y^3=$

We can try to find a common term to better organize the exercise and we will obtain:

$16(X^4\cdot4Y^3)$

**Perfect! Now, let's move on to a complex and slightly different example:**

**Power of a power advanced examples:**

**$(2^{X+3})^X\cdot(2^X)^4=$**

Don't worry, even if there are mathematical operations among the exponents, the properties do not change.

Let's start with the first expression which is a bit more complex. We learned that, when we have a power of a power we multiply the exponents.

We will multiply the entire exponent that is inside the parentheses by the entire exponent located outside the parentheses. We will do the same with the other term and we will obtain:

$2^{(X+3)\cdot X}\cdot2^{4X}=$

We will multiply the exponents of the first expression and we will obtain:

$2^{X^2+3X}\cdot2^{4X}=$

Now let's remember that, if we have a multiplication operation between equal bases we can add the exponents.

**We will do this and we will obtain:**

$2^{X^2+3X+4X}=$

We simplify terms in the exponent and it will give us:

$2^{X^2+7X}=$

**Simplify the following expression:**

$\frac{\left(3x^3y^2\right)^2\left(2x^2y^4\right)^4}{\left(2xy^2\right)^3}$

To simplify the expression, first apply the power of a product property, which allows us to raise each of the factors inside the parenthesis to the indicated power, then apply the power of a power property. We obtain:

$\frac{\left(3^2\left(x^3\right)^2\left(y^2\right)^2\right)\cdot\left(2^4\left(x^2\right)^4\left(y^4\right)^4\right)}{2^3\left(x^{}\right)^3\left(y^2\right)^3}=\frac{\left(9x^6y^4\right)\cdot\left(16x^8y^{16}\right)}{8x^3y^6}$

Finally, apply the properties of products and quotients of powers with the same base:

$\frac{144x^{6+8}y^{4+16}}{8x^3y^6}=\frac{114x^{14}y^{20}}{8x^3y^6}=18x^{14-3}y^{20-6}=18x^{11}y^{14}$

**If you are interested in this article, you might also be interested in the following articles:**

- Powers
- The Rules of Exponentiation
- Division of Powers with the Same Base
- Power of a Multiplication
- Power of a Quotient
- Power with Zero Exponent
- Powers with a Negative Integer Exponent
- Taking Advantage of All the Properties of Powers or Laws of Exponents
- Exponentiation of Whole Numbers

**In the** **Tutorela** **blog, you will find a variety of articles about mathematics.**

$\left(4^2\right)^2=$

$\left(3^3\right)^2=$

$\left(2^2\right)^2=$

$\left(5^2\right)^5=$

$\left(7^2\right)^2=$

$\left(X^{2-4}\right)^2=$

$\left(X^{2+4}\right)^3=$

$\left(X^{21-7}\right)^2=$

$\left(X^{11-9}\right)^3=$

$\left(X^{5-3}\right)^3=$

$(\frac{4X^5}{X})^2\cdot(\frac{3Y^3}{Y})^3=$

$(\frac{4Y^5}{2Y})^2\cdot(\frac{Y^4}{2Y})^4=$

$(\frac{2X^5}{2X})^3\cdot(\frac{X^3}{2X})^3=$

$(\frac{2Y^5}{2Y^5})^3\cdot(\frac{X^3}{2X^3})^3=$

$(\frac{2Y^5}{2Y^5})^3\cdot(\frac{X^3}{2X^3})^3\cdot(\frac{2Y^3}{2Y^6})^2\cdot(\frac{X^2}{2X^2})^2=$

$(3^{X+7})^X\cdot(3^X)^3=$

$(2^{X-2})^X\cdot(3^{X-2})^6=$

$(3^{X-3})^X\cdot(3^{X-3})^3=$

$(3^{2-3})^2\cdot(7^{X-5})^2=$

$(8^{X-3})^X\cdot(72^{X+2})^X=$

Test your knowledge

Question 1

\( [(4-2)^2]^3= \)

**What is a power of a power?**

A power of a power is an expression in which we raise a power to another power.

**What is a power of a power and example?**

A power of a power is a power in which the base is also a power, for example:

- $\left(3^2\right)^5$
- $\left(5^{2x+1}\right)^2$

**How do you calculate a power of a power?**

To solve a power of a power, we must multiply the exponents, and the result of the multiplication is placed as the exponent on the initial base.

**Assignment**

$2^3\times2^4+(4^3)^2+\frac{2^5}{2^3}=$

**Solution**

$2^3\cdot2^4+\left(4^3\right)^2+\frac{2^5}{2^3}=$

$2^{3+4}+4^{3\cdot2}+2^{\left(5-3\right)}=$

$2^7+4^6+2^2$

**Answer**

$2^2+2^7+4^6$

Do you know what the answer is?

Question 1

\( [(4-2)^2]^3= \)

**Assignment**

$(4^x)^y=$

**Solution**

$\left(4^x\right)^y=4^{x\cdot y}$

**Answer**

$4^{xy}$

**Assignment**

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

**Solution**

We will use the formula

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

$2^{2\cdot3}+3^{3\cdot4}\cdot9^{2\cdot6}=$

$2^6+3^{12}\cdot9^{12}$

**Answer**

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

Check your understanding

Question 1

\( [(4-2)^2]^3= \)

**Assignment**

$(4^2)^3+(g^3)^4=$

**Solution**

We will use the formula

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

$4^{2\cdot3}+9^{3\cdot4}=$

$4^6+9^{12}$

**Answer**

$4^6+9^{12}$

**Assignment**

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

**Solution**

We will use the formula

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

$(7\cdot3)^{2\cdot6}+3^{-1\cdot3}\cdot2^{3\cdot4}=$

$21^{12}+3^{-3}\cdot2^{12}$

**Answer**

$21^{12}+3^{-3}\cdot2^{12}$

Do you think you will be able to solve it?

Question 1

\( [(4-2)^2]^3= \)