Multiplication of Powers: Using multiple rules

To determine which value is greater, let's simplify each choice:

Choice 1: $(a^2)^4$
By using the power of a power rule: $(x^m)^n = x^{m \times n}$, it simplifies to:
$(a^2)^4 = a^{2 \times 4} = a^8$.

Choice 2: $a^2 + a^0$
Evaluate using the zero exponent rule, $a^0 = 1$:
This expression becomes $a^2 + 1$.

Choice 3: $a^2 \times a^1$
Apply the product of powers rule: $x^m \times x^n = x^{m+n}$:
This simplifies to $a^{2+1} = a^3$.

Choice 4: $\frac{a^{14}}{a^9}$
Apply the quotient of powers rule: $\frac{x^m}{x^n} = x^{m-n}$:
This simplifies to $a^{14-9} = a^5$.

Now, let's compare these simplified forms:
We have $a^8$, $a^2 + 1$, $a^3$, and $a^5$.

For $a > 1$, exponential functions grow rapidly, thus:
- $a^8$ is greater than $a^5$.
- $a^8$ is greater than $a^3$.
- $a^8$ is greater than $a^2 + 1$ for sufficiently large $a$.

Thus, the expression with the highest power, and therefore the greatest value, is $(a^2)^4$.

Let's first deal with the first term in the multiplication, noting that the terms in the numerator and denominator have identical bases, so we'll use the power rule for division between terms with the same base:

$\frac{a^m}{a^n}=a^{m-n}$We'll apply for the first term in the expression:

$\frac{a^{3b}}{a^{2b}}\cdot a^b=a^{3b-2b}\cdot a^b=a^b\cdot a^b$where we also simplified the expression we got as a result of subtracting the exponents of the first term,

Next, we'll notice that the two terms in the multiplication have identical bases, so we'll use the power rule for multiplication between terms with identical bases:

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

$a^b\cdot a^b=a^{b+b}=a^{2b}$Therefore, the correct answer is A.

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

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

In both the numerator and denominator, multiplication occurs between terms with identical bases, so we'll apply the power law for multiplying terms with identical bases:

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

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, instead we will place terms next to each other.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

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

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

Note that division is required between two terms with identical bases, hence we'll apply the power law for division between terms with identical bases:

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

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:

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

In the final step we calculated the subtraction between the exponents.

This is the most simplified form of the expression:

Therefore, the correct answer is C.

To solve this problem, we need to simplify and compare the given expressions.

Let's simplify each:

• $y^7 \times y^2$:
  Using the product of powers rule, $y^7 \times y^2 = y^{7+2} = y^9$.
• $(y^4)^3$:
  Using the power of a power rule, $(y^4)^3 = y^{4 \times 3} = y^{12}$.
• $y^9$:
  This is already in its simplest form, $y^9$.
• $\frac{y^{11}}{y^4}$:
  Using the power of a quotient rule, $\frac{y^{11}}{y^4} = y^{11-4} = y^7$.

Now that all the expressions are in the form $y^n$, we can compare the exponents to see which is greatest: $y^9$, $y^{12}$, $y^9$, and $y^7$.

The expression with the highest power is $y^{12}$, which corresponds to the choice $(y^4)^3$.

Thus, the greater value among the choices is $(y^4)^3$.

We need to calculate division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we will use the law of exponents for division between terms with identical 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.

In this problem, there is also a term with a negative exponent, but this does not pose an issue regarding the use of the aforementioned law of exponents. In fact, this law of exponents is valid in all cases for numerical terms with different exponents, including negative exponents, rational number exponents, and even irrational number exponents, etc.

Let's return to the problem and apply the aforementioned law of exponents for each fraction separately:

$\frac{y^3}{y^6}\cdot\frac{y^4}{y^{-2}}\cdot\frac{y^{12}}{y^7}=y^{3-6}\cdot y^{4-(-2)}\cdot y^{12-7}=y^{-3}\cdot y^6\cdot y^5$

When in the second stage we applied the aforementioned law of exponents for the second fraction (from left to right) carefully, this is because the term in the denominator of this fraction has a negative exponent and according to the aforementioned law of exponents, we need to subtract between the exponent of the numerator and the exponent of the denominator, which in this case gave us subtraction of a negative number from another number, an operation we performed carefully.

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

Let's return to the problem and note that we need to perform multiplication between terms with identical bases, therefore we will use the law of exponents for multiplication between terms with identical base:

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

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

Let's apply this law in the problem:

$y^{-3}y^6y^5=y^{-3+6+5}=y^8$

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

Therefore the correct answer is B.

We want to calculate each of the expressions separately, however - in order to do this more efficiently, we will first deal with the multiplication terms between the roots in both expressions separately:

a. Let's start with the left expression, the multiplication of roots in this expression is:

$\sqrt{10}\cdot\sqrt{10}$

We'll apply the laws of exponents in order to simplify this expression, noting that the expression is actually multiplying the number by itself and therefore can be written as a term to the second power:

$\sqrt{10}\cdot\sqrt{10}=(\sqrt{10})^2$

Now let's recall the definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

And the law of exponents for power to power:

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

Let's apply these two laws and calculate the value of the above expression:

$(\sqrt{10})^2 =(10^{\frac{1}{2}})^2=10^{\frac{1}{2}\cdot2}=10^1=10$

Where in the first step we converted the root in parentheses to a half power using the definition of root as a power mentioned earlier, and in the next step we applied the law of power to power that was also mentioned earlier, then we simplified the expression.

b. Let's continue to the multiplication of roots in the right expression:

$\sqrt{5}\cdot\sqrt{20}$

In addition to the definition of root as a power mentioned earlier, let's also recall the law of exponents for powers in parentheses where terms are multiplied but in the opposite direction:

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

The literal interpretation of this law in the direction given here is that a multiplication between two terms with equal power exponents can be written as a multiplication between the bases in parentheses raised to that same power,

Let's return to the expression in question and apply both laws of exponents mentioned:

$\sqrt{5}\cdot\sqrt{20} =5^{\frac{1}{2}}\cdot20^{\frac{1}{2}}=(5\cdot20)^{\frac{1}{2}}$

Where in the first step we converted the roots to half powers using the definition of root as a power, and in the next step we applied the last mentioned law of exponents in its specified direction, since both terms in the multiplication here have the same power,

Let's continue and simplify the expression we got:

$(5\cdot20)^{\frac{1}{2}} =100^{\frac{1}{2}}=\sqrt{100}=10$

Where in the first step we calculated the value of the multiplication in parentheses, in the next step we returned to writing roots using the definition of root as power, but in the opposite direction, in the final step we calculated the numerical value of the root,

Let's summarize a and b above, we got that:

$\sqrt{10}\cdot\sqrt{10}=(\sqrt{10})^2 =10$ and

$\sqrt{5}\cdot\sqrt{20} =(5\cdot20)^{\frac{1}{2}} =\sqrt{100}=10$,

Let's return to the original problem and use this information:

$3^2+\sqrt{10}\cdot\sqrt{10}\text{ }_{\textcolor{red}{—}\text{ }}2^3+\sqrt{5}\cdot\sqrt{20}:5 \\ \downarrow\\ 3^2+10\text{ }_{\textcolor{red}{—}\text{ }}2^3+10:5$

Let's continue and handle both expressions together, in the left expression we'll first calculate the value of the term in the power and then the result of the addition,

And in the right expression we'll first calculate the result of the term in the power and the result of the division operation and add between the results:

$3^2+10\text{ }_{\textcolor{red}{—}\text{ }}2^3+10:5 \\ \downarrow\\ 9+10\text{ }_{\textcolor{red}{—}\text{ }}8+2 \\ \downarrow\\ 19\text{ }_{\textcolor{red}{—}\text{ }}10$

Therefore the left expression gives a higher result, meaning the trend between the expressions is:

$19>10$

Therefore the correct answer is answer B.

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.

Let's deal with the first term in the problem, which is the fraction,

For this, we'll recall two laws of exponents:

a. The law of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$b. The law of exponents for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$Let's apply these laws of exponents to the problem:

$$$\frac{17^{-3}\cdot17^{3x}}{17}-17x=\frac{17^{-3+3x}}{17}-17x=17^{-3+3x-1}-17x=17^{3x-4}-17x$where in the first stage we'll apply the law of exponents mentioned in 'a' above to the fraction's numerator, and in the next stage we'll apply the law of exponents mentioned in 'b' to the resulting expression, then we'll simplify the expression.

Therefore, the correct answer is answer a.

Write the problem in an organized way using fraction notation for the first term:$\frac{}{}\frac{X^3\cdot X^2}{X^5}+X^4$

Let's continue and refer to the first term in the above sum:

$\frac{X^3\cdot X^2}{X^5}$

Begin with the numerator, using the law of exponents for multiplying terms with identical bases:

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

and we obtain the following:

$\frac{X^3\cdot X^2}{X^5}=\frac{X^{3+2}}{X^5}=\frac{X^5}{X^5}$

Now proceed to use the law of exponents for the division between terms with identical bases:

$a^m:a^n=\frac{a^m}{a^n}=a^{m-n}$

When in the first stage of the above formula we just wrote the same thing in fraction notation instead of using division (:), let's apply the law of exponents to the problem and calculate the result for the first term that we obtained above:

$\frac{X^5}{X^5}=X^{5-5}=X^0$

Proceed to apply the law of exponents:

$a^0=1$

Note that this rule is actually just the understanding that dividing a number by itself will always give the result 1. Let's return to the problem and we obtain the result of the first term in the exercise (meaning - the result of calculating the fraction) is:

$X^0=1$,

Let's return to the complete exercise and summarize everything said so far as follows:

$\frac{X^3\cdot X^2}{X^5}+X^4=\frac{X^5}{X^5}+X^4=X^0+X^4=1+X^4$

To solve this problem, we need to simplify the given expression using the rules of exponents:

First, simplify inside the brackets:
$\frac{a^4}{a^3} \times \frac{a^8}{a^7} = a^{4-3} \times a^{8-7} = a^1 \times a^1 = a^{1+1} = a^2$

Now, handle the entire expression, dividing it by $\frac{a^{10}}{a^8}$:
$\frac{a^2}{\frac{a^{10}}{a^8}} = a^2 \times \frac{a^8}{a^{10}} = a^2 \times a^{8-10} = a^2 \times a^{-2} = a^{2 + (-2)} = a^0$

Recall that any non-zero number raised to the power of zero is 1, hence: $a^0 = 1$

Therefore, the solution to the problem is $1$.

First, simplify the numerator and the denominator separately:
Numerator: $X^4 \cdot X^3 = X^{4+3} = X^7$
Denominator: $X^5 \cdot X^2 = X^{5+2} = X^7$

Now, combine the simplified numerator and denominator:

$\frac{X^7}{X^7}$

Since any number divided by itself is 1, we have:

$\frac{X^7}{X^7} = 1$

Therefore, the correct answer is:

$1$

First we will rearrange the expression and use the fact that multiplying a fraction means multiplying the numerator of the fraction, and the distributive property of multiplication:

$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=x^3\cdot x^4\cdot 2\cdot \frac{1}{x^3}\cdot x^{-8}=2\cdot x^3\cdot x^4\cdot \frac{1}{x^3}\cdot x^{-8}$$$Next, we'll use the law of exponents for negative exponents:

$a^{-n}=\frac{1}{a^n}$We'll apply the law of exponents to the expression in the problem:

$2\cdot x^3\cdot x^4\cdot\frac{1}{x^3}\cdot x^{-8} =2\cdot x^3\cdot x^4\cdot x^{-3}\cdot x^{-8}$When we applied the above law of exponents for the fraction in the multiplication,

From now on, we will no longer use the multiplication sign and will switch to the conventional notation where juxtaposition of terms means multiplication between them,

Now we'll recall the law of exponents for multiplying terms with the same base:

$a^m\cdot a^n=a^{m+n}$And we'll apply this law of exponents to the expression we got in the last step:

$2x^3x^4x^{-3}x^{-8} =2x^{3+4+(-3)+(-8)} =2x^{3+4-3-8}=2x^{-4}$When in the first stage we applied the above law of exponents and in the following stages we simplified the expression in the exponent,

Let's summarize the solution steps so far, we got that:
$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}= 2 x^3x^4 x^{-3} x^{-8} =2x^{-4}$

Now let's note that there is no such answer in the given options, a further check of what we've done so far will also reveal that there is no calculation error,

Therefore, we can conclude that additional mathematical manipulation is required to determine which is the correct answer among the suggested answers,

Let's note that in answers A and B there are similar expressions to the one we got in the last stage, however - we can directly rule out the other two options since they are clearly different from the expression we got,

Furthermore, we'll note that in the expression we got, x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator (and both are in the numerator - Note at the end of the solution on this topic), so we'll rule out this answer,

If so - we are left with only one option - which is answer A', however we want to verify (and need to verify!) that this is indeed the correct answer:

Let's note that in the expression we got x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator , which reminds us of the law of exponents for negative exponents mentioned at the beginning of the solution,

In addition, let's note that in answer B x is in the second power but inside parentheses that are also in the second power, whereas in the expression we got in the last stage of solving the problem x is in the fourth power which might remind us of the law of exponents for power to a power,

We'll check this, starting with the law of exponents for negative exponents mentioned at the beginning of the solution, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Next, we'll represent the term with the negative exponent that we got in the last stage of solving the problem, as a term in the denominator of the fraction with a positive exponent:

$2x^{-4}=2\cdot\frac{1}{x^4}$When we applied the above law of exponents,

Next, let's note that using the law of exponents for power to a power, but in the opposite direction:

$a^{m\cdot n}= (a^m)^n$We can conclude that:

$x^4=x^{2\cdot2}=(x^2)^2$Therefore, we'll return to the expression we got in the last stage and apply this understanding:

$2\cdot\frac{1}{x^4} =2\cdot\frac{1}{(x^2)^2}$ Let's summarize then the problem-solving stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4} =2\cdot\frac{1}{(x^2)^2}$Let's note that we still haven't got the exact expression suggested in answer A, but we are already very close,

To reach the exact expression claimed in answer A, we'll recall another important law of exponents, and a useful mathematical fact:

Let's recall the law of exponents for exponents applying to terms in parentheses, but in the opposite direction:

$\frac{a^n}{c^n}=\big(\frac{a}{c}\big)^n$And let's also recall the fact that raising the number 1 to any power will yield the result 1:

$1^{x}=1$And therefore we can write the expression we got in the last stage in the following way:

$2\cdot\frac{1}{(x^2)^2}=2\cdot\frac{1^2}{(x^2)^2}$And then since in the numerator and denominator of the fraction there are terms with the same exponent we can apply the above law of exponents, and represent the fraction whose numerator and denominator are terms with the same exponent as a fraction whose numerator and denominator are the bases of the terms and it is raised to the same exponent:

$2\cdot\frac{1^2}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$ Let's summarize then the solution stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4}=2\cdot\frac{1}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$And therefore the correct answer is indeed answer A.

Note:

When it's written "the number in the numerator" despite the fact that there is no fraction in the expression at all, it's because we can always refer to any number as a number in the numerator of a fraction if we remember that any number divided by 1 equals itself, that is, we can always write a number as a fraction by writing it like this:

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

Apply the power law for multiplication between terms with identical bases:

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

$\big(-\frac{1}{8}\big)^8\cdot\big(-\frac{1}{8}\big)^{-3}=\big(-\frac{1}{8}\big)^{8+(-3)}=\big(-\frac{1}{8}\big)^{8-3}=\big(-\frac{1}{8}\big)^5$

In the first stage we applied the above power law and in the following stages we simplified the expression in the exponent,

Let's continue and use the power law for power of terms in parentheses:

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

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

$\big(-\frac{1}{8}\big)^5=\big(-1\cdot\frac{1}{8}\big)^5=(-1)^5\cdot\big(\frac{1}{8}\big)^5=-1\cdot\big(\frac{1}{8}\big)^5=-\big(\frac{1}{8}\big)^5$

In the first stage we presented the expression in parentheses as a multiplication between negative one and a positive number. In the next stage we applied the above power law and then simplified the expression we obtained whilst noting that negative one to an odd power will (always) give the result negative one.

Next we'll recall two additional power laws:

a. The negative power law:

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

b. The power law for power of a power:

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

We'll continue and apply these two laws to the expression that we obtained in the last stage:

$-\big(\frac{1}{8}\big)^5=-(8^{-1})^5=-8^{(-1)\cdot5}=-8^{-5}$

In the first stage we presented the fraction inside the parentheses as a term with a negative power using the above power law for negative power mentioned in a. above. In the next stage we applied the power law for power of a power mentioned in b. above carefully, given that the term inside the parentheses has a negative power. We then simplified the expression in the exponent.

Let's summarize the solution :

$\big(-\frac{1}{8}\big)^8\cdot\big(-\frac{1}{8}\big)^{-3}=\big(-\frac{1}{8}\big)^5=-\big(\frac{1}{8}\big)^5=-(8^{-1})^5=-8^{-5}$

Therefore the correct answer is answer d.

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

$\frac{a^{20b}}{a^{15b}}\cdot\frac{a^{3b}}{a^{2b}}=\frac{a^{20b}\cdot a^{3b}}{a^{15b}\cdot a^{2b}}$

In both the numerator and denominator, multiplication occurs between terms with identical bases, thus we'll apply 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 this point forward, we will no longer use the multiplication sign, instead we will place terms next to each other.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{a^{20b}a^{3b}}{a^{15b}a^{2b}}=\frac{a^{20b+3b}}{a^{15b+2b}}=\frac{a^{23b}}{a^{17b}}$

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

Now we need to perform division between two terms with identical bases, thus we'll apply the power law for dividing terms with identical bases:

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

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:

$\frac{a^{23b}}{a^{17b}}=a^{23b-17b}=a^{6b}$

In the final step we calculate the subtraction between the exponents.

This is the most simplified form of the expression:

Therefore, the correct answer is D.

We'll begin by applying the multiplication law between fractions, 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, instead we will place the terms next to each other.

Note that in both the numerator and denominator, multiplication is performed between terms with identical bases, therefore we'll apply 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 the multiplication in the numerator and denominator:

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

In the last step we calculated the sum of the exponents.

Now we need to perform division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we'll apply 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:

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

In the last step we calculated the result of the subtraction operation in the exponent.

We cannot simplify the expression further. Therefore the correct answer is D.

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 will now apply the law of negative exponents, but in reverse:

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

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 that we obtained 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 obtained the following,

Let's summarize the solution so far as shown below:

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

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 that we obtained 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 as follows:

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

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.

Proceed to solve this in two stages. In the first stage, we'll use the power rule for powers in parentheses:

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

which states that when a power is applied to terms in parentheses, it applies to each term inside the parentheses when they are opened,

Let's apply this rule to our problem:

$\big((x^{\frac{1}{4}}\cdot3^2\cdot6^3)^{\frac{1}{4}}\big)^8=((x^{\frac{1}{4}})^{\frac{1}{4}}\cdot(3^2)^{\frac{1}{4}}\cdot(6^3)^{\frac{1}{4}})^8$

When opening the parentheses, we applied the power to each term separately, however given that each of these terms is raised to a power, we did this carefully and used parentheses,

Next, we'll use the power rule for a power raised to a power:

$(b^m)^n=b^{m\cdot n}$

Let's apply this rule to the expression that we obtained:

$(x^{\frac{1}{4}\cdot\frac{1}{4}}\cdot3^{2\cdot\frac{1}{4}}\cdot6^{3\cdot\frac{1}{4}})^8=(x^{\frac{1}{16}}\cdot3^{\frac{2}{4}}\cdot6^{\frac{3}{4}})^8=x^{\frac{1}{16}\cdot8}\cdot3^{\frac{2}{4}\cdot8}\cdot6^{\frac{3}{4}\cdot8}=x^{\frac{8}{16}}\cdot3^{\frac{16}{4}}\cdot6^{\frac{24}{4}}$

In the second stage we performed multiplication in the fractions of the power expressions of the terms that we obtained. Remember that multiplication in fractions is actually multiplication in the numerator. In the final stage we simplified the fractions in the power expressions of the multiplication terms that we obtained:

$x^{\frac{8}{16}}\cdot3^{\frac{16}{4}}\cdot6^{\frac{24}{4}}=x^{\frac{1}{2}}\cdot3^4\cdot6^6$

Therefore, the correct answer is answer B.

To determine which expression has the greatest value, we apply the exponent rules to simplify each choice:

• For $x^3 \times x^4$, using the product rule: $x^3 \times x^4 = x^{3+4} = x^7$.
• For $(x^3)^5$, using the power of a power rule: $(x^3)^5 = x^{3 \times 5} = x^{15}$.
• $x^{10}$ is already in its simplest form.
• For $\frac{x^9}{x^2}$, using the quotient rule: $\frac{x^9}{x^2} = x^{9-2} = x^7$.

To identify the greater value, we compare the exponents:

• $x^7$ from choices 1 and 4.
• $x^{15}$ from choice 2.
• $x^{10}$ from choice 3.

The expression with the largest exponent is $(x^3)^5$ or $x^{15}$.

Therefore, the expression with the greatest value is $(x^3)^5$.