Power of a Quotient Rule for Exponents: Using multiple rules

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.

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:

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

Next, we'll note 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 this point forward, we will no longer use the multiplication sign, but instead use the conventional notation where placing terms next to each other implies multiplication.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

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

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

Now we notice that we need to perform division between two terms with identical bases, so we'll use the power law for dividing 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:

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

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 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.

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, let's write the problem in an organized way and use 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:

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

Let's deal with the numerator, first using the law of exponents for multiplying terms with identical bases:

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

and we get:

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

Now let's use the law of exponents for 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 we got above:

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

Now let's use the law of exponents:

$a^0=1$

We can notice 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 get that 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, we got:

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

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.

First we'll use 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:

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

where 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 we got in the last stage:

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

where 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 got while 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 we got in the last stage:

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

where 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, since the term inside the parentheses has a negative power and then simplified the expression in the exponent.

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

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

Therefore the correct answer is answer d.

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.

Let's start with multiplying the two fractions in the problem using the rule for fraction multiplication, which states that we multiply numerator by numerator and denominator by denominator while keeping the fraction line:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

But before we perform the fraction multiplication, let's note

Important Note-

Notice that in both fractions of the multiplication there are expressions that are binomials squared

Of course, we're referring to-$(5+3x)^8$and $(5+3x)^6$,

We'll treat these expressions simply as terms squared - meaning:

We won't open the parentheses and we'll treat both expressions exactly as we treat terms:

$a^8$and$a^6$.

Now let's return to the problem and continue from where we left off:

$$Let's apply the rule for fraction multiplication mentioned above in the problem and perform the multiplication between the fractions:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}=\frac{136xy^5(5+3x)^8}{3xy^2y(5+3x)^6}=\frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6}$

where in the first stage we performed the multiplication between the fractions using the above rule, then we simplified the expression in the fraction's denominator using the distributive property of multiplication, and the law of exponents for multiplying terms with identical bases:

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

which we applied in the final stage to the fraction's denominator that we got.

Now we'll use the above rule for fraction multiplication again, but in the opposite direction in order to present the resulting fraction as a multiplication of fractions where each fraction contains only numbers or terms with identical bases:

$\frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6}=\frac{136}{3}\cdot\frac{x}{x}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}\cdot1\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}$

where additionally in the second stage we applied the fact that dividing any number by itself gives 1.

We can now continue and simplify the expression using the law of exponents for division between terms with identical bases:

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

Let's apply the above law to the last expression we got:

$\frac{136}{3}\cdot\frac{y^5}{y^3}\cdot\frac{(5+3x)^8}{(5+3x)^6}=\frac{136}{3}y^{5-3}(5+3x)^{8-6}=\frac{136}{3}y^2(5+3x)^2=45\frac{1}{3}\cdot y^2(5+3x)^2$

where in the first stage we applied the above law of exponents, treating the binomial as noted above, and then simplified the resulting expression, in the final stage we converted the improper fraction we got to a mixed number,

Let's summarize the solution to the problem, we got that:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}= \frac{136xy^5(5+3x)^8}{3xy^3(5+3x)^6} =45\frac{1}{3}\cdot y^2(5+3x)^2$

Therefore the correct answer is answer B.

Another Important Note:

In solving the problem above we detailed the steps to the solution, and used fraction multiplication in both directions multiple times and the above law of exponents,

We could have shortened the process, applied the distributive property of multiplication, and performed directly both the application of the above law of exponents and the numerical reduction to get directly the last line we got:

$\frac{136xy^5}{3xy^2}\cdot\frac{(5+3x)^8}{(5+3x)^6\cdot y}=45\frac{1}{3}\cdot y^2(5+3x)^2$

(meaning we could have skipped the part where we presented the fraction as a multiplication of fractions and even the initial fraction multiplication we performed and gone straight to reducing the fractions)

However, it should be emphasized that this quick solution method is conditional on the fact that between all terms in the numerator and denominator of each fraction in the problem, and also between the fractions themselves, multiplication is performed, meaning, that we can put it into a single fraction line like we did at the beginning and we can apply the distributive property and present as fraction multiplication as above, etc., this is a point worth noting, since not every problem we encounter will meet all the conditions mentioned here in this note.

Let's start with multiplying the two fractions in the problem using the rule for multiplying fractions, which states that we multiply numerator by numerator and denominator by denominator while keeping the fraction line:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

But before we perform the multiplication of fractions, let's note

Important note-

Notice that in both fractions in the multiplication there are expressions that are binomials squared

Of course, we're referring to-$(5x+4y)^3$ and$(5x+4y)^2$,

We'll treat these expressions simply as terms squared - meaning:

We won't open the parentheses and we'll treat both expressions exactly as we treat terms:

$a^3$and$a^2$.

Now let's return to the problem and continue from where we left off:

$$Let's apply the above-mentioned rule for multiplying fractions in the problem and perform the multiplication between the fractions:

$\frac{(5x+4y)^3\cdot7x}{45y\cdot x^2}\cdot\frac{3xy}{(5x+4y)^2}=\frac{7\cdot3xxy(5x+4y)^3}{45\cdot x^2y(5x+4y)^2}=\frac{7x^2y(5x+4y)^3}{15x^2y(5x+4y)^2}$

where in the first stage we performed the multiplication between the fractions using the above rule, and in the second stage we reduced the numerical part in the resulting fraction, then we simplified the expressions in the numerator and denominator of the resulting fraction using the distributive property of multiplication and the law of exponents for multiplying terms with identical bases:

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

which we applied in the final stage to the numerator and denominator of the resulting fraction.

Now we'll use the above rule for multiplying fractions again, but in the opposite direction in order to express the resulting fraction as a multiplication of fractions so that each fraction contains only numbers or terms with identical bases:

$\frac{7x^2y(5x+4y)^3}{15x^2y(5x+4y)^2}=\frac{7}{15}\cdot\frac{x^2}{x^2}\cdot\frac{y}{y}\cdot\frac{(5x+4y)^3}{(5x+4y)^2}=\frac{7}{15}\cdot1\cdot1\cdot\frac{(5x+4y)^3}{(5x+4y)^2}=\frac{7}{15}\cdot\frac{(5x+4y)^3}{(5x+4y)^2}$

where additionally in the second stage we applied the fact that dividing any number by itself gives 1.

We can now continue and simplify the expression using the law of exponents for division between terms with identical bases:

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

Let's apply the above law to the last expression we got:

$\frac{7}{15}\cdot\frac{(5x+4y)^3}{(5x+4y)^2}=\frac{7}{15}\cdot(5x+4y)^{3-2}=\frac{7}{15}\cdot(5x+4y)^{1}=\frac{7}{15}\cdot(5x+4y)$

where in the first stage we applied the above law of exponents, treating the binomial as noted above, and then simplified the resulting expression remembering that raising a number to the power of 1 gives the number itself,

Let's summarize the solution to the problem, we got that:

$\frac{(5x+4y)^3\cdot7x}{45y\cdot x^2}\cdot\frac{3xy}{(5x+4y)^2}= \frac{7}{15}\cdot\frac{(5x+4y)^3}{(5x+4y)^2} =\frac{7}{15}\cdot(5x+4y)$

Therefore the correct answer is answer D.

Another important note:

In solving the problem above we detailed the steps to the solution, and used fraction multiplication in both directions multiple times and the above law of exponents,

We could have shortened the process, applied the distributive property of multiplication, and performed directly both the application of the above law of exponents and the reduction of the numerical part to get directly the last line we got:

$\frac{(5x+4y)^3\cdot7x}{45y\cdot x^2}\cdot\frac{3xy}{(5x+4y)^2}=\frac{7}{15}\cdot(5x+4y)$

(meaning we could have skipped the part where we expressed the fraction as a multiplication of fractions and even the initial multiplication of fractions we performed and gone straight to reducing between the fractions)

However, it should be emphasized that this quick solution method is conditional on the fact that between all terms in the numerator and denominator of each of the fractions in the problem, and also between the fractions themselves, multiplication is performed, meaning, that we can combine into one unified fraction as we did at the beginning and can apply the distributive property of multiplication and express as multiplication of fractions as mentioned above, etc., this is a point worth noting, since not in every problem we encounter all the conditions mentioned here in this note are met.

We use the three appropriate power properties to solve the problem:

1. Power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$ 2. Power law for an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$ 3. Power law for the division of terms with identical bases:

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

We continue and apply the three previous laws to the problem:

$2^3\cdot2^4+(4^3)^2+\frac{2^5}{2^3}=2^{3+4}+4^{3\cdot2}+2^{5-3}=2^7+4^6+2^2$

In the first step we apply the power law mentioned in point 1 to the first expression on the left, the power law mentioned in point 2 to the second expression on the left, and the power law mentioned in point 3 to the third expression on the left, separately. In the second step, we simplify the expressions by exponents possession of the received terms,

Then,after using the substitution property for addition, we find that the correct answer is D.

In order to simplify the given expression, we will use the following two laws of exponents:

a. Law of exponents for multiplication of terms with identical bases:

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

b. Law of exponents for division of terms with identical bases:

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

Let's solve the given expression:

$\frac{2^3\cdot2^4}{2^5}=$First, since in the numerator we have multiplication of terms with identical bases, we'll use the law of exponents mentioned in a:

$\frac{2^3\cdot2^4}{2^5}= \\ \frac{2^{3+4}}{2^5}=\\ \frac{2^{7}}{2^5}=\\$We'll continue, since we have division of terms with identical bases, we'll use the law of exponents mentioned in b:

$\frac{2^{7}}{2^5}=\\ 2^{7-5}=\\ 2^2=\\ \boxed{4}$

Let's summarize the simplification of the given expression:

$\frac{2^3\cdot2^4}{2^5}= \\ \frac{2^{7}}{2^5}=\\ 2^2=\\ \boxed{4}$

Therefore, the correct answer is answer d.

In order to solve the problem we must use two power laws, as shown below:

A. Power property for terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$B. Power property for an exponent raised to another exponent:

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

We will apply these two power laws to the problem in two steps:

Let's start by applying the power law specified in A to the second term from the left in the given problem:

$\frac{2^7}{2^4}=2^{7-4}=2^3$In the first step we apply the power law specified in A and then proceed to simplify the resulting expression,

We then advance to the next step and apply the power law specified in B to the third term from the left in the given problem :

$(8^2)^5=8^{2\cdot5}=8^{10}$In the first stage we apply the power law specified in B and then proceed to simplify the resulting expression,

Let's summarize the two steps listed above to solve the general problem:

$(4\cdot7)^9+\frac{2^7}{2^4}+(8^2)^5= (4\cdot7)^9+2^3+8^{10}$In the final step, we calculate the result of multiplying the terms within the parentheses in the first term from the left:

$(4\cdot7)^9+2^3+8^{10}=28^9+2^3+8^{10}$Therefore, the correct answer is option c.

In solving this problem we will use three laws of exponents, let's recall them:

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

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

b. The law of exponents for power of a power:

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

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

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

We will apply these three laws of exponents to the expression in the problem in three stages:

Let's start by applying the law of exponents mentioned in a to the first term from the left in the expression:

$4^7\cdot4^{-1}= 4^{7+(-1)} = 4^{7-1}=4^{6}$

When in the first stage we applied the law of exponents mentioned in a and in the following stages we simplified the resulting expression,

We'll continue to the next stage and apply the law of exponents mentioned in b and deal with the second term from the left in the expression:

$(3^2)^7=3^{2\cdot7}=3^{14}$

When in the first stage we applied the law of exponents mentioned in b and in the following stages we simplified the resulting expression,

We'll continue to the next stage and apply the law of exponents mentioned in c and deal with the third term from the left in the expression:

$\frac{9^5}{9^2} =9^{5-2}=9^3$

When in the first stage we applied the law of exponents mentioned in c and in the following stages we simplified the resulting expression,

Let's summarize the three stages detailed above for the complete solution of the problem:

$4^7\cdot4^{-1}+(3^2)^7+\frac{9^5}{9^2} =4^6+3^{14}+9^3$

Therefore the correct answer is answer c.

The problem involves multiplication between two fractions, so first we'll apply the rule for multiplying fractions which states that multiplication between two fractions is calculated by putting one fraction over a line by multiplying the numerators together and multiplying the denominators together, mathematically:

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

Let's apply it to the problem:

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

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

$a^m\cdot a^n=a^{m+n}$Let's apply it separately to the numerator and denominator in the problem:

$\frac{a^{2x}\cdot a^{2y}}{a^y\cdot a^{-y}}=\frac{a^{2x+2y}}{a^{y+(-y)}}=\frac{a^{2x+2y}}{a^0}$

Next, we'll remember that any number to the power of 0 equals 1, mathematically:

$a^0=1$

So let's return to the problem:

$\frac{a^{2x+2y}}{a^{0}}=\frac{a^{2x+2y}}{1}=a^{2x+2y}$

Where we actually used the fact that division by 1 doesn't change the value of the number, meaning mathematically:

$X:1=\frac{X}{1}=X$

Now let's examine the result we got above:

$a^{2x+2y}\text{ }$

In terms of simplification using the laws of exponents we have indeed finished since this is the most simplified expression,

but it's worth noting that in the exponent we got an expression that can be factored using common factor extraction:

$2x+2y=2(x+y)$

In this case the common factor is the number 2,

Let's return to the result of the expression simplification, we got:

$a^{2x+2y}=a^{2(x+y)}$Therefore the correct answer is C.

a. We'll start by handling the first expression from the left, which is a multiplication of fractions. However, we won't perform the fraction multiplication; instead, we'll notice that in each of the fractions in the multiplication, there are terms in both numerator and denominator with identical bases,

therefore we'll use the power law for division between terms with identical bases:

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

We'll apply this to our problem in the first expression from the left and apply the above power law separately for each term in the fraction multiplication:

$\frac{a^7}{a^5}\cdot\frac{a^{-2}}{a^{-4}}=a^{7-5}\cdot a^{-2-(-4)}=a^2\cdot a^{-2+4}=a^2\cdot a^2$

Next, we'll notice that we got an algebraic expression that is actually the term multiplied by itself, therefore from the definition of power we can write the expression as a power:

$a^2\cdot a^2=(a^2)^2$

(Of course we could also apply the multiplication law between terms with identical bases to the above expression, but here we want to also recall the power of a power law, so we'll choose this way),

Now let's recall the power of a power law:

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

and apply it to the expression we got:

$(a^2)^2=a^{2\cdot2}=a^4$

We've finished handling the first expression from the left, meaning we got that:

$\frac{a^7}{a^5}\cdot\frac{a^{-2}}{a^{-4}}=a^4$

We'll remember this result and move on to the next expression.

b. For the second expression from the left, we'll apply again the power of a power law mentioned in section a:

$(a^7)^3=a^{7\cdot3}=a^{21}$

and we've finished handling this expression as well.

Let's summarize the two solution sections a and b into the simplified expression result:

$\frac{a^7}{a^5}\cdot\frac{a^{-2}}{a^{-4}}+(a^7)^3=a^4+a^{21}$

Therefore the correct answer is b.

Let us begin by addressing the given problem as an equation:

$\big( \big(\frac{1}{5} \big)^2 \big)^?:5=125$Therefore, we shall replace the question mark with an x and proceed to solve it:

$\big( \big(\frac{1}{5} \big)^2 \big)^x:5=125$ Remember that dividing by a certain number is equivalent to multiplying by its inverse, so we will rewrite the given equation bearing this in mind:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$

Let's briefly discuss the solution technique:

Generally speaking the goal when solving exponential equations is to achieve a situation where there is a term on each of the two sides of the equation so that both sides have the same base. In such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and thus solve a simple equation for the unknown.

Mathematically, we will perform a mathematical manipulation (according to the laws of equation manipulation) on both sides of the equation. Or we will concentrate on the development of one of the sides of the equation with the help of power rules and algebra in order to reach the following situation:

$b^{m(x)}=b^{n(x)}$when $m(x),\hspace{4pt}n(x)$Algebraic expressions ( functions of the unknown $x$) that can also exclude the unknowns ($x$) that we are trying to find in the problem, which is the solution to the equation,

It is then stated that:

$m(x)=n(x)$and we solve the simple equation that we obtained.

We return to solving the equation in the given problem:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$In solving this equation, various power rules are used:

a. Power property with negative exponent:

$a^{-n}=\frac{1}{a^n}$b. Power property for a power of an exponent raised to another exponent:

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

Our initial goal is to simplify the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)

To do this, we will start by dealing with the fraction on the left side of the equation:

$\frac{1}{5}$That is, both the fraction inside the parenthesis and the fraction outside the parenthesis, this is done with the help of the power rule for a negative exponent specified in A above. We then represent this fraction as a term with a negative power and in the next step we apply the power rule for a power of an exponent raised to another exponent specified in B above. We then are able to remove the parentheses starting from the inner parenthesis to the outer ones. This is shown below step by step:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ (5^{(-1)\cdot 2} )^x\cdot 5^{-1}=125 \\ 5^{(-1)\cdot 2\cdot x} \cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\$When we carry out the development of the left side of the equation as described above, we initially apply the power rule for a negative exponent mentioned above in A.

In the following steps we apply the power rule for a power of an exponent raised to another exponent as mentioned above in B. We remove the parentheses: starting from the inner parenthesis to the outer. In the last step we simplify the expression in the power exponent on the left side of the equation,

c. Later we remember the power property for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Thus we apply this law to the left side of the equation that we obtained in the last step.

$5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x+(-1)}=125 \\ 5^{-2x-1}=125 \\$In the first step we apply the aforementioned power law to the product between members with identical bases mentioned above in C and in the following steps we simplify the expression in the power exponent on the left side,

Next, we seek to obtain the same base on both sides of the equation, the best way to achieve this is by decomposing each of the numbers in the problem into prime factors (using powers as well), you will notice that the number 125 is a power of the number 5, that is:

$125=5^3$This is the presentation (factorization) of the number 125 using its prime factor, which is the number 5.

So we return to the equation we obtained in the previous step and replace this number with its decomposition into prime factors:

$5^{-2x-1}=125 \\ 5^{-2x-1}=5^3 \\$We have reached our goal, we have obtained an equation in which both sides have terms with the same base, therefore we can state that the power exponents of the terms on both sides are equal, and in order solve the resulting equation for the unknown, we proceed as follows:

$5^{-2x-1}=5^3 \\ \\ \downarrow\\ -2x-1=3$We will continue to solve the resulting equation by isolating the unknown on the left side. We can achieve this in an usual way, by moving the sections and dividing the final equation by the unknown's coefficient:

$-2x-1=3 \\ -2x=3+1\\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \frac{\not{-2}x}{\not{-2}}=\frac{4}{-2}\\ x=-\frac{4}{2}\\ \bm{x=-2 }$In the first step we simplify the equation by moving the sides, remembering that when a term is moved its sign changes, then we complete the isolation by nullifying dividing both sides of the equation by its coefficient. In the last steps, we simplify the expression obtained by reducing the fractions,

We have thus solved the given equation. Below is a brief step by step summary of the solution:

$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x-1}=5^3 \\ \downarrow\\ -2x-1=3 \\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \bm{x=-2 }$Therefore, the correct answer is option a.