Applying Combined Exponents Rules: Single Variable

We use the formula:

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

and therefore we obtain:

$(a^5)^7=a^{5\times7}=a^{35}$

First, we'll enter the same fraction using the multiplication law between fractions, by multiplying numerator by numerator and denominator by denominator:

$\frac{x}{y}\cdot\frac{w}{z}=\frac{x\cdot w}{y\cdot z}$

Let's return to the problem and apply the above law:

$\frac{a^{12}}{a^9}\cdot\frac{a^3}{a^4}=\frac{a^{12}\cdot a^3}{a^{^9}\cdot a^4}$

From here on we will no longer indicate the multiplication sign, but use the conventional writing method where placing terms next to each other means multiplication.

Now we'll notice that both in the numerator and denominator, multiplication is performed between terms with identical bases, therefore we'll use the power law for multiplication between terms with the same base:

$b^m\cdot b^n=b^{m+n}$

Note that this law can only be used to calculate multiplication between terms with identical bases.

Let's return to the problem and calculate separately the results of multiplication in the numerator and denominator:

$\frac{a^{12}a^3}{a^{^9}a^4}=\frac{a^{12+3}}{a^{9+4}}=\frac{a^{15}}{a^{13}}$

where in the last step we calculated the sum of the exponents.

Now, we'll notice that we need to perform division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we'll use the power law for division between terms with the same base:

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

Note that this law can only be used to calculate division between terms with identical bases.

Let's return to the problem and apply the above law:

$\frac{a^{15}}{a^{13}}=a^{15-13}=a^2$

where in the last step we calculated the result of subtraction in the exponent.

We got the most simplified expression possible and therefore we're done,

therefore the correct answer is D.

Let's start with multiplying the fractions, remembering that multiplication of fractions is performed by multiplying numerator by numerator and denominator by denominator:

$\frac{b^{22}}{b^{20}}\cdot\frac{b^{30}}{b^{20}}=\frac{b^{22}\cdot b^{30}}{b^{20}\cdot b^{20}}$

Next, we'll notice that both in the numerator and denominator, multiplication occurs between terms with identical bases, so we'll use the power law for multiplying terms with identical bases:

$c^m\cdot c^n=c^{m+n}$

We emphasize that this law can only be used when multiplication is performed between terms with identical bases.

From here on, we will no longer indicate the multiplication sign, but use the conventional writing method where placing terms next to each other means multiplication.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{b^{22}b^{30}}{b^{20}b^{20}}=\frac{b^{22+30}}{b^{20+20}}=\frac{b^{52}}{b^{40}}$

where in the final step we calculated the sum of the exponents in the numerator and denominator.

Now we notice that division is required between two terms with identical bases, so we'll use the power law for division between terms with identical bases:

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

We emphasize that this law can only be used when division is performed between terms with identical bases.

Let's return to the problem and apply the above power law:

$\frac{b^{52}}{b^{40}}=b^{52-40}=b^{12}$

where in the final step we calculated the subtraction between the exponents.

We have obtained the most simplified expression and therefore we are done.

Therefore, the correct answer is C.

First we rewrite the first expression on the left of the problem as a fraction:

$\frac{a^2}{a}+a^3\cdot a^5$Then we use two properties of exponentiation, to multiply and divide terms with identical bases:

1.

$b^m\cdot b^n=b^{m+n}$

2.

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

Returning to the problem and applying the two properties of exponentiation mentioned earlier:

$\frac{a^2}{a}+a^3\cdot a^5=a^{2-1}+a^{3+5}=a^1+a^8=a+a^8$

Later on, keep in mind that we need to factor the expression we obtained in the last step by extracting the common factor,

Therefore, we extract from outside the parentheses the greatest common divisor to the two terms which are:

$a$ We obtain the expression:

$a+a^8=a(1+a^7)$ when we use the property of exponentiation mentioned earlier in A.

$$$a^8=a^{1+7}=a^1\cdot a^7=a\cdot a^7$

Summarizing the solution to the problem and all the steps, we obtained the following:

$\frac{a^2}{a}+a^3\cdot a^5=a(1+a^{7)}$

Therefore, the correct answer is option b.

We use the formula:

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

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

We use the formula:

$(a\times b)^n=a^nb^n$

$(5\times x\times3)^3=(15x)^3$

$(15x)^3=(15\times x)^3$

$15^3x^3$

We use the formula

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

Therefore, we obtain:

$a^{4\times6}=a^{24}$

We use the formula

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

Therefore, we obtain:

$((b^3)^6)^2=(b^{3\times6})^2=(b^{18})^2=b^{18\times2}=b^{36}$

Let us begin by using the law of exponents for a power that is applied to parentheses in which terms are multiplied:

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

We apply the rule to our problem:

$(x\cdot4\cdot3)^3= x^3\cdot4^3\cdot3^3$

When we apply the power to the product of the terms within parentheses, we apply the power to each term of the product separately and keep the product,

Therefore, the correct answer is option C.

We use the power rule of distributing exponents.

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

$\big((y^6)^8\big)^9=(y^{6\cdot8})^9=y^{6\cdot8\cdot9}=y^{432}$When we use the aforementioned rule twice, the first time for the inner parentheses in the first stage and the second time for the remaining parentheses in the second stage, in the last stage we calculate the result of the multiplication in the power exponent.

Therefore, the correct answer is option b.

We use the power law for multiplication within parentheses:

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

We apply it in the problem:

$(y\cdot7\cdot3)^4=y^4\cdot7^4\cdot3^4$

Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we can understand that it applies not only to two terms within parentheses, but also for multiple terms within parentheses.

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

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

$Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2$When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

We begin by using the negative exponent rule.

$b^{-n}=\frac{1}{b^n}$We apply it to the problem:

$$$$$a^{-4}=\frac{1}{a^4}$Therefore, the correct answer is option B.

We use the power rule for exponents.

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

$\big((a^2)^3\big)^{\frac{1}{4}}=(a^{2\cdot3})^{\frac{1}{4}}=a^{2\cdot3\cdot\frac{1}{4}}=a^{\frac{6}{4}}=a^{\frac{3}{2}}$When we use the previously mentioned rule twice, the first time for the inner parentheses in the first stage and the second time for the remaining parentheses in the second stage, in the third stage we calculate the result of the multiplication in the exponent. While remembering that multiplying by a fraction is actually doubling the numerator of the fraction and, finally, in the last stage we simplify the fraction we obtained in the exponent.

Now remember that -

$\frac{3}{2}=1\frac{1}{2}=1.5$

Therefore, the correct answer is option a.