Applying Combined Exponents Rules: Variables in the exponent of the power

We begin by using the distributive law of exponents.

$(x\cdot y)^n=x^n\cdot y^n$

We apply this property to the given problem :

$(2\cdot3\cdot7\cdot9)^{ab+3} =2^{ab+3}3^{ab+3}7^{ab+3}9^{ab+3}$

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.

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

$(x\cdot y)^n=x^n\cdot y^n$

We apply this law to the problem expression:

$(5\cdot12\cdot4\cdot6)^{a+3bx}=5^{a+3bx}12^{a+3bx}4^{a+3bx}6^{a+3bx}$

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.

We use the power law for a multiplication between parentheses:

$(z\cdot t)^n=z^n\cdot t^n$

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:

$(4\cdot9\cdot11)^a=4^a\cdot9^a\cdot11^a$

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.

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

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

$((14^{3x})^{2y})^{5a}=(14^{3x})^{2y\cdot5a}=14^{3x\cdot2y\cdot5a}=14^{30xya}$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.

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

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

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)

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

$(2\cdot4\cdot8)^{a+3}= 2^{a+3}4^{a+3}8^{a+3}$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.

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

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

$((3^9)^{4x})^{5y}= (3^9)^{4x\cdot 5y} =3^{9\cdot4x\cdot 5y}=3^{180xy}$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.

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

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

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

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

Therefore, the correct answer is option c.

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

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

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

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

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

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

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

Therefore, the correct answer is option d.

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

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

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

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

$7^x\cdot7^{-x}=7^{x+(-x)}=7^{x-x}=7^0$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:

$X^0=1$We obtain:

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

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

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:

$X^0=1$

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

$\frac{b^m}{b^n}=b^{m-n}$

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 $a$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 $a\neq0$ we can claim that:

$1=a^0$

Let's apply this to the problem:

$\frac{1}{a^n}=\frac{a^0}{a^n}$

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:

$\frac{a^0}{a^n}=a^{0-n}=a^{-n}$

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

$\frac{1}{a^n}=\frac{a^0}{a^n}=a^{-n}$

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

We'll use the power rule for powers:

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

$((4x)^{3y})^2= (4x)^{3y\cdot2}=(4x)^{6y}$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:

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

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

Therefore, the correct answer is answer D.