Examples with solutions for Applying Combined Exponents Rules: 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

4a×9a×11a 4^a\times9^a\times11^a

Exercise #2

(4x)y= (4^x)^y=

Video Solution

Step-by-Step Solution

Using the law of powers for an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n} We apply it in the problem:

(4x)y=4xy (4^x)^y=4^{xy} Therefore, the correct answer is option a.

Answer

4xy 4^{xy}

Exercise #3

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}

Exercise #4

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 #5

(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 #6

((39)4x)5y= ((3^9)^{4x)^{5y}}=

Video Solution

Step-by-Step Solution

We use the power rule for an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n} We apply this rule to the given problem:

((39)4x)5y=(39)4x5y=394x5y=3180xy ((3^9)^{4x})^{5y}= (3^9)^{4x\cdot 5y} =3^{9\cdot4x\cdot 5y}=3^{180xy} In the first step we applied the previously mentioned power rule and removed the outer parentheses. In the next step we applied the power rule once again and removed the remaining parentheses. In the final step we simplified the resulting expression.

Therefore, the correct answer is option b.

Answer

3180xy 3^{180xy}

Exercise #7

((143x)2y)5a= ((14^{3x})^{2y})^{5a}=

Video Solution

Step-by-Step Solution

Using the power rule for an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n} We apply the rule to the given problem:

((143x)2y)5a=(143x)2y5a=143x2y5a=1430xya ((14^{3x})^{2y})^{5a}=(14^{3x})^{2y\cdot5a}=14^{3x\cdot2y\cdot5a}=14^{30xya} In the first step we applied the aforementioned power rule and removed the outer parentheses. In the next step we again applied the power rule and removed the remaining parentheses.

In the final step we simplified the resulting expression,

Therefore, through the rule of substitution (which is applied to the exponent of the power in the obtained expression) it can be concluded that the correct answer is answer D.

Answer

1430axy 14^{30axy}

Exercise #8

22x+12523x= 2^{2x+1}\cdot2^5\cdot2^{3x}=

Video Solution

Step-by-Step Solution

Since the bases are the same, the exponents can be added:

2x+1+5+3x=5x+6 2x+1+5+3x=5x+6

Answer

25x+6 2^{5x+6}

Exercise #9

42y454y46= 4^{2y}\cdot4^{-5}\cdot4^{-y}\cdot4^6=

Video Solution

Step-by-Step Solution

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

aman=am+n a^m\cdot a^n=a^{m+n} We apply the property for this problem:

42y454y46=42y+(5)+(y)+6=42y5y+6 4^{2y}\cdot4^{-5}\cdot4^{-y}\cdot4^6= 4^{2y+(-5)+(-y)+6}=4^{2y-5-y+6} We simplify the expression we got in the last step:

42y5y+6=4y+1 4^{2y-5-y+6} =4^{y+1} When we add similar terms in the exponent.

Therefore, the correct answer is option c.

Answer

4y+1 4^{y+1}

Exercise #10

72x+1717x= 7^{2x+1}\cdot7^{-1}\cdot7^x=

Video Solution

Step-by-Step Solution

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

aman=am+n a^m\cdot a^n=a^{m+n} We apply the property to our expression:

72x+1717x=72x+1+(1)+x=72x+11+x 7^{2x+1}\cdot7^{-1}\cdot7^x=7^{2x+1+(-1)+x}=7^{2x+1-1+x} We simplify the expression we got in the last step:

72x+11+x=73x 7^{2x+1-1+x}=7^{3x} When we add similar terms in the exponent.

Therefore, the correct answer is option d.

Answer

73x 7^{3x}

Exercise #11

Insert the corresponding expression:

(15)xy= \left(15\right)^{xy}=

Video Solution

Step-by-Step Solution

To solve this problem, we will rewrite the expression (15)xy (15)^{xy} using the rules of exponents.

  • Step 1: Understand that (15)xy (15)^{xy} can be rewritten using the power of a power rule.
  • Step 2: Apply the exponent rule (am)n=am×n(a^m)^n = a^{m \times n}. We know (15x)y=(15)x×y(15^x)^y = (15)^{x \times y} and (15y)x=(15)y×x(15^y)^x = (15)^{y \times x}, both equivalent to (15)xy (15)^{xy} .
  • Step 3: Analyze each choice:

Choice 1: (15y)x (15^y)^x is equivalent to (15)xy(15)^{xy} since applying the rule gives us (15y)x=(15)y×x=(15)xy(15^y)^x = (15)^{y \times x} = (15)^{xy}.
Choice 2: (15x)y (15^x)^y is also equivalent to (15)xy(15)^{xy} because applying the rule provides (15x)y=(15)x×y=(15)xy(15^x)^y = (15)^{x \times y} = (15)^{xy}.
Choice 3: 15x×15y 15^x \times 15^y results in 15x+y15^{x+y}, which is not equivalent to (15)xy(15)^{xy} as it uses the product of powers rule.\
Choice 4: Both (15y)x (15^y)^x and (15x)y (15^x)^y are correct based on the rules involved.

Based on the analysis, choice 4 (a'+b' are correct) is the correct answer.
Both (15y)x(15^y)^x and (15x)y(15^x)^y are equivalent representations of (15)xy (15)^{xy}.

Answer

a'+b' are correct

Exercise #12

Insert the corresponding expression:

103x= 10^{3x}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Understand the expression 103x10^{3x}.
  • Step 2: Apply the power of a power rule to rewrite it.
  • Step 3: Identify the correct equivalent expression from the options.

Now, let's work through each step:
Step 1: The expression given is 103x10^{3x}, which involves a base of 10 and a combination of numerical and variable exponents, specifically 3x3x.
Step 2: To rewrite this expression, we use the power of a power rule for exponents, which states (am)n=amn(a^m)^n = a^{m \cdot n}. In our case, we want to reverse this process: we express 103x10^{3x} as (103)x(10^3)^x. Here, by viewing 3x3x as the product of 33 and xx, we can apply the rule effectively.
Step 3: We now compare our converted expression (103)x(10^3)^x with the provided answer choices. The correct rewritten form is:
- Choice 3: (103)x\left(10^3\right)^x
Therefore, the solution to the problem is (103)x\left(10^3\right)^x. This matches the correct answer provided, validating our analysis and application of the power of a power rule.

Answer

(103)x \left(10^3\right)^x

Exercise #13

7x7x=? 7^x\cdot7^{-x}=\text{?}

Video Solution

Step-by-Step Solution

We use the law of exponents to multiply terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n} We apply the law to given the problem:

7x7x=7x+(x)=7xx=70 7^x\cdot7^{-x}=7^{x+(-x)}=7^{x-x}=7^0 In the first stage we apply the above power rule and in the following stages we simplify the expression obtained in the exponent,

Subsequently, we use the zero power rule:

X0=1 X^0=1 We obtain:

70=1 7^0=1 Lastly we summarize the solution to the problem.

7x7x=7xx=70=1 7^x\cdot7^{-x}=7^{x-x}=7^0 =1 Therefore, the correct answer is option B.

Answer

1 1

Exercise #14

3x2x32x= 3^x\cdot2^x\cdot3^{2x}=

Video Solution

Step-by-Step Solution

In this case we have 2 different bases, so we will add what can be added, that is, the exponents of 3 3

3x2x32x=2x33x 3^x\cdot2^x\cdot3^{2x}=2^x\cdot3^{3x}

Answer

33x2x 3^{3x}\cdot2^x

Exercise #15

1an=? \frac{1}{a^n}=\text{?}

a0 a\ne0

Video Solution

Step-by-Step Solution

This question is actually a proof of the law of exponents for negative exponents, we will prove it simply using two other laws of exponents:

a. The zero exponent law, which states that raising any number to the power of 0 (except 0) will give the result 1:

X0=1 X^0=1

b. The law of exponents for division between terms with identical bases:

bmbn=bmn \frac{b^m}{b^n}=b^{m-n}

Let's return to the problem and pay attention to two things, the first is that in the denominator of the fraction there is a term with base a a and the second thing is that according to the zero exponent law mentioned above in a' we can always write the number 1 as any number (except 0) to the power of 0, particularly in this problem, given that a0 a\neq0 we can claim that:

1=a0 1=a^0

Let's apply this to the problem:

1an=a0an \frac{1}{a^n}=\frac{a^0}{a^n}

Now that we have in the numerator and denominator of the fraction terms with identical bases, we can use the law of division between terms with identical bases mentioned in b' in the problem:

a0an=a0n=an \frac{a^0}{a^n}=a^{0-n}=a^{-n}

Let's summarize the steps above, we got that:

1an=a0an=an \frac{1}{a^n}=\frac{a^0}{a^n}=a^{-n}

In other words, we proved the law of exponents for negative exponents and understood why the correct answer is answer c.

Answer

an a^{-n}

Exercise #16

((4x)3y)2= ((4x)^{3y})^2=

Video Solution

Step-by-Step Solution

We'll use the power rule for powers:

(am)n=amn (a^m)^n=a^{m\cdot n} We'll apply this rule to the expression in the problem:

((4x)3y)2=(4x)3y2=(4x)6y ((4x)^{3y})^2= (4x)^{3y\cdot2}=(4x)^{6y} When in the first stage we applied the mentioned power rule and eliminated the outer parentheses, in the next stage we simplified the resulting expression,

Next, we'll recall the power rule for powers that applies to parentheses containing a product of terms:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n We'll apply this rule to the expression we got in the last stage:

(4x)6y=46yx6y (4x)^{6y} =4^{6y}\cdot x^{6y} When we applied the power to the parentheses to each term of the product inside the parentheses.

Therefore, the correct answer is answer D.

Answer

46yx6y 4^{6y}\cdot x^{6y}