Examples with solutions for Power of a Product: Variables in the exponent of the power

Exercise #1

Solve the following exercise:

(4×9×11)a (4\times9\times11)^a

Video Solution

Step-by-Step Solution

We use the power law for a multiplication between parentheses:

(zt)n=zntn (z\cdot t)^n=z^n\cdot t^n That is, a power applied to a multiplication between parentheses is applied to each term when the parentheses are opened,

We apply it in the problem:

(4911)a=4a9a11a (4\cdot9\cdot11)^a=4^a\cdot9^a\cdot11^a Therefore, the correct answer is option b.

Note:

From the power property formula mentioned, we can understand that it works not only with two terms of the multiplication between parentheses, but also valid with a multiplication between multiple terms in parentheses. As we can see in this problem.

Answer

4a9a11a 4^a9^a11^a

Exercise #2

Simplify:

(51246)a+3bx (5\cdot12\cdot4\cdot6)^{a+3bx}

Video Solution

Step-by-Step Solution

Use the power property for a power in parentheses where there is a multiplication of its terms:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n We apply this law to the problem expression:

(51246)a+3bx=5a+3bx12a+3bx4a+3bx6a+3bx (5\cdot12\cdot4\cdot6)^{a+3bx}=5^{a+3bx}12^{a+3bx}4^{a+3bx}6^{a+3bx} When we apply a power to parentheses where its terms are multiplied, we do it separately and keep the multiplication.

Therefore, the correct answer is option d.

Answer

5a+3bx12a+3bx4a+3bx6a+3bx 5^{a+3bx}12^{a+3bx}4^{a+3bx}6^{a+3bx}

Exercise #3

(248)a+3= (2\cdot4\cdot8)^{a+3}=

Video Solution

Step-by-Step Solution

Let's begin by using the distributing exponents rule (An exponent outside of a parentheses needs to be distributed across all the numbers and variables within the parentheses)

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n We first apply this rule to the given problem:

(248)a+3=2a+34a+38a+3 (2\cdot4\cdot8)^{a+3}= 2^{a+3}4^{a+3}8^{a+3} When then we apply the power to each of the terms of the product inside the parentheses separately and maintain the multiplication.

The correct answer is option d.

Answer

2a+34a+38a+3 2^{a+3}4^{a+3}8^{a+3}

Exercise #4

Simplify:

(2379)ab+3 (2\cdot3\cdot7\cdot9)^{ab+3}

Video Solution

Step-by-Step Solution

We begin by using the distributive law of exponents.

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n We apply this property to the given problem :

(2379)ab+3=2ab+33ab+37ab+39ab+3 (2\cdot3\cdot7\cdot9)^{ab+3} =2^{ab+3}3^{ab+3}7^{ab+3}9^{ab+3} When we apply the power of parentheses to each of the terms of the product inside the parentheses separately and maintain the multiplication.

Therefore, the correct answer is option a.

Answer

2ab+33ab+37ab+39ab+3 2^{ab+3}3^{ab+3}7^{ab+3}9^{ab+3}