Applying Combined Exponents Rules: Using laws of exponents with parameters

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.

Let's recall the law of exponents for multiplication between terms with identical bases:

$b^m\cdot b^n=b^{m+n}$We'll apply this law to the fraction in the expression in the problem:

$\frac{a^4a^8a^{-7}}{a^9}=\frac{a^{4+8+(-7)}}{a^{^9}}=\frac{a^{4+8-7}}{a^9}=\frac{a^5}{a^9}$where in the first stage we'll apply the aforementioned law of exponents and in the following stages we'll simplify the resulting expression,

Let's now recall the law of exponents for division between terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We'll apply this law to the expression we got in the last stage:

$\frac{a^5}{a^9}=a^{5-9}=a^{-4}$Let's now recall the law of exponents for negative exponents:

$b^{-n}=\frac{1}{b^n}$And we'll apply this law of exponents to the expression we got in the last stage:

$a^{-4}=\frac{1}{a^4}$Let's summarize the solution steps so far, we got that:

$\frac{a^4a^8a^{-7}}{a^9}=\frac{a^5}{a^9}=\frac{1}{a^4}$Therefore, the correct answer is answer A.

To solve the given problem, we'll apply the laws of exponents to simplify the expression $\frac{a^b b^a}{c^b} \cdot b^{-c} \cdot \frac{1}{a}$.

Let's go through each step:

• Start with the expression $\frac{a^b b^a}{c^b} \cdot b^{-c} \cdot \frac{1}{a}$.
• Rewrite $b^{-c}$ as $\frac{1}{b^c}$ using the rule $x^{-n} = \frac{1}{x^n}$.
• Substitute it back: $\frac{a^b b^a}{c^b} \cdot \frac{1}{b^c} \cdot \frac{1}{a}$.
• Combine the expressions into a single fraction: $\frac{a^b b^a}{c^b \cdot b^c \cdot a}$.
• Use the rule $x^m \cdot x^n = x^{m+n}$ to simplify the exponents in the numerator and denominator: - In the numerator, no changes necessary since terms are already separated. - In the denominator, combine $b^a$ and $b^c$ using the exponent rule: $b^{a+c}$.
• Update the fraction: $\frac{a^b}{a} \cdot \frac{b^a}{b^{a+c}} \cdot \frac{1}{c^b}$.
• Simplify each component: - $\frac{a^b}{a} = a^{b-1}$, - $\frac{b^a}{b^{a+c}} = b^{a-(a+c)} = b^{-c} = \frac{1}{b^c}$.
• Combine all components: $\frac{a^{b-1}}{c^b \cdot b^c}$.
• Express the result combining all simplified terms: $\frac{1}{a^{1-b} b^{c-a} c^b}$.

Therefore, the solution to the problem is $\frac{1}{a^{1-b} b^{c-a} c^b}$.

Let's begin by dealing with the root in the problem. We'll use the root and exponent law for this:

$\sqrt[n]{a^m}=(\sqrt[n]{a})^m=a^{\frac{m}{n}}$

Apply the above exponent law to the problem:

$\frac{1}{x^7}\cdot y^7\cdot\sqrt[4]{x^8}=\frac{1}{x^7}\cdot y^7\cdot x^{\frac{8}{4}}=\frac{1}{x^7}\cdot y^7\cdot x^2$

When in the first stage we applied the above law to the third term in the product. We did this carefully whilst paying attention to what goes into the numerator of the fraction in the exponent. Let's ask ourselves what goes into the denominator of the fraction in the exponent? In the following stages, we simplified the expression that we obtained.

Next, we'll recall the exponent law for negative exponents in the opposite direction:

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

We'll apply this exponent law to the first term in the product in the expression that we obtained in the last stage:

$\frac{1}{x^7}\cdot y^7\cdot x^2=x^{-7}\cdot y^7\cdot x^2=y^7\cdot x^{-7}\cdot x^2$

When in the first stage we applied the above exponent law to the first term in the product and in the next stage we arranged the expression that we obtained by using the commutative property of multiplication. Hence terms with identical bases are adjacent to each other.

Next, we'll recall the exponent law for multiplying terms with identical bases:

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

Apply this exponent law to the expression that we obtained in the last stage:

$y^7\cdot x^{-7}\cdot x^2=y^7x^{-7+2}=y^7x^{-5}$

When in the first stage we applied the above exponent law for the terms with identical bases, and then proceeded to simplify the expression that we obtained. Additionally in the final stages we removed the · sign and switched to the conventional notation where placing terms next to each other signifies multiplication.

Let's summarize the various steps of the solution so far:

$\frac{1}{x^7}\cdot y^7\cdot\sqrt[4]{x^8}=\frac{1}{x^7}\cdot y^7\cdot x^{\frac{8}{4}}=x^{-7}y^7x^2=y^7x^{-5}$

Therefore, the correct answer is answer D.