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.

$$We'll use the law of exponents for negative exponents, but in the opposite direction:

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

We'll apply this law to the problem for the third term in the product:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n}\cdot n^{-m}\cdot m^{-1}=m^{-n}\cdot m^{-1}\cdot n^{-m}$

When in the first stage we applied the above law of exponents for the third term in the product, and in the next stage we rearranged the resulting expression using the distributive property of multiplication so that terms with identical bases are adjacent to each other,

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

$a^x\cdot a^y=a^{x+y}$

And we'll apply this law of exponents to the expression we got in the last stage:

$m^{-n}\cdot m^{-1}\cdot n^{-m}=m^{-n+(-1)}\cdot n^{-m}=m^{-n-1}\cdot n^{-m}$

When in the first stage we applied the above law of exponents for terms with identical bases, and in the next stage we simplified the expression with the exponent of the first term in the product in the expression we got,

Let's summarize the solution so far, we got that:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n}\cdot n^{-m}\cdot m^{-1}=m^{-n-1}\cdot n^{-m}$

Now let's note that there is no such answer among the given options, and an additional check of what we've done so far will reveal that there is no calculation error,

Therefore, we can conclude that additional mathematical manipulation is needed to determine which is the correct answer among the given options,

Let's note that options B and D have expressions similar to the expression we got in the last stage, while the other two options can be directly eliminated since they are clearly different from the expression we got,

Furthermore, let's note that in addition, the second term in the product in the expression we got, which is the term-$n^{-m}$, is in the numerator (note at the end of the solution on this topic), while in option B it's in the denominator, so we'll eliminate this option,

Thus - we're left with only one option - which is answer D, however we want to verify (and must verify!) that this is indeed the correct answer:

We'll do this using the law of exponents for negative exponents that we mentioned earlier, but in the forward direction:

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

And we'll deal separately with the first term in the product in the expression we got in the last stage of solving the problem, which is the term:

$m^{-n-1}$

Let's note that we can represent the expression in the exponent as follows:

$-n-1=-(n+1)$

Where we used factoring out and took out negative one from the parentheses,

Next, we'll use the above law of exponents and the last understanding to represent the above expression (which we're currently dealing with, separately) as a term in the denominator of a fraction:

$m^{-n-1}=m^{-\underline {\bm{(n+1)}}}=\frac{1}{m^{\underline {\bm{n+1}}}}$

When in the first stage, in order to use the above law of exponents - we represented the term in question as having a negative exponent, while using the fact that:

$-n-1=-(n+1)$,

Next, we applied the above law of exponents carefully, since the number that- x represents in our use of the above law of exponents here is:

$n+1$(underlined in the expression above)

Let's return then to the expression we got in the last stage of solving the given problem, and apply for the first term in the product the mathematical manipulation we just performed:

$m^{-n-1}\cdot n^{-m}=m^{-(n+1)}\cdot n^{-m}=\frac{1}{m^{n+1}}\cdot n^{-m}$

Now let's simplify the expression we got and perform the multiplication in the fraction while remembering that multiplication in a fraction means multiplying the numerators:

$\frac{1}{m^{n+1}}\cdot n^{-m}=\frac{1\cdot n^{-m}}{m^{n+1}}=\frac{n^{-m}}{m^{n+1}}$

Let's summarize then the solution stages so far, we got that:

$m^{-n}\cdot n^{-m}\cdot\frac{1}{m}=m^{-n-1}\cdot n^{-m} =\frac{n^{-m}}{m^{n+1}}$

Therefore, the correct answer is indeed answer D.

Note:

When it's written "the number in the numerator" even though there isn't actually a fraction in the expression, this is because we can always refer to any number as being in the numerator of a fraction if we remember that any number divided by 1 equals itself, meaning, we can always write a number as a fraction like this:

$X=\frac{X}{1}$and therefore we can actually refer to $X$as a number in the numerator of a fraction.

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.