Applying Combined Exponents Rules: Using multiple rules

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.

$(\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}$

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 handle each expression in the problem separately:

a. We'll start with the leftmost expression, first calculating the result of the multiplication in parentheses, and then use the power rule for power to a power:

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

Let's apply this to the problem for the first expression from the left:

$((7\cdot3)^2)^6=(21^2)^6=21^{2\cdot6}=21^{12}$

where in the final step we calculated the result of multiplication in the power expression,

We're done with this expression, let's move on to the next expression from the left.

b. Let's continue with the second expression from the left, using the power rule for power to a power that we mentioned above and apply it separately to each factor in this expression:

$(3^{-1})^3\cdot(2^3)^4=3^{-1\cdot3}\cdot2^{3\cdot4}=3^{-3}\cdot2^{12}$

Note that the multiplication factors we got have different bases, so we cannot further simplify this expression,

Therefore, let's combine parts a and b above in the result of the original problem:

$((7\cdot3)^2)^6+(3^{-1})^3\cdot(2^3)^4=21^{12}+3^{-3}\cdot2^{12}$

Therefore, the correct answer is answer d.

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.

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

$(\frac{1}{3})^3=\frac{1^3}{3^3}$

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

First we'll use the fact that raising any number to the power of 0 gives the result 1, mathematically:

$X^0=1$

We'll apply this to both the numerator and denominator of the fraction in the problem:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1\cdot6^7}{36^4\cdot1}=\frac{6^7}{36^4}$

Next we'll note that -36 is a power of the number 6:

$36=6^2$

And we'll use this fact in the denominator to get expressions with identical bases in both the numerator and denominator:

$\frac{6^7}{36^4}=\frac{6^7}{(6^2)^4}$

Now we'll recall the power rule for power of a power to simplify the expression in the denominator:

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

And we'll also recall the power rule for division between terms with identical bases:

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

We'll apply these two rules to the expression we got above:

$\frac{6^7}{(6^2)^4}=\frac{6^7}{6^{2\cdot4}}=\frac{6^7}{6^8}=6^{7-8}=6^{-1}$

Where in the first stage we applied the first rule we mentioned earlier - the power of a power rule and simplified the expression in the exponent of the denominator term, then in the next stage we applied the second power rule mentioned before - the division rule for terms with identical bases, and again simplified the expression in the resulting exponent,

Finally we'll use the power rule for negative exponents:

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

And we'll apply it to the expression we got:

$6^{-1}=\frac{1}{6}$

Let's summarize everything we did, we got that:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1}{6}$

Therefore the correct answer is A.

Let's start by handling the term in the multiplication that is in parentheses:

$(-y)^3$

For this, we'll recall the law of exponents for an exponent of a term in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

Accordingly, we get that:

$(-y)^3=(-1\cdot y)^3=(-1)^3\cdot y^3=-1\cdot y^3=-y^3$

We'll use this understanding in the problem and apply it to the aforementioned term:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=\frac{y^3\cdot y^{-4}\cdot(-y^3)}{y^{-3}}=\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}}$

where in the first stage we used the above understanding carefully - while using parentheses, and this is in order to remember that we're dealing with multiplication (not subtraction) and then we rearranged the expression using the distributive property of multiplication while remembering that a negative coefficient means multiplying by negative one,

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

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

and we'll apply this law to the expression we got in the last stage:

$\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}}=\frac{-y^{3+(-4)+3}}{y^{-3}}=\frac{-y^{3-4+3}}{y^{-3}}=-\frac{y^2}{y^{-3}}$

where in the first stage we applied the above law of exponents to the multiplication terms (with identical bases) in the expression and in the final stage we remembered that negative one divided by negative one equals negative one.

Let's summarize the solution steps so far:

$\frac{y^3\cdot y^{-4}\cdot(-y)^3}{y^{-3}}=\frac{-y^3\cdot y^{-4}\cdot y^3}{y^{-3}} =-\frac{y^2}{y^{-3}}$

We'll continue and recall the law of exponents for dividing terms with identical bases:

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

Let's apply this law to the expression we got in the last stage:

$-\frac{y^2}{y^{-3}}=-y^{2-(-3)}=-y^{2+3}=-y^5$

where in the first stage we applied the above law of exponents carefully, because the term in the denominator has a negative exponent and then we simplified the expression in the exponent,

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

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

Therefore, the correct answer is answer A.

Note:

Let's note and emphasize that the minus sign in the final answer is not under the exponent, meaning - the exponent doesn't apply to it but only to $y$, and this is in contrast to the understanding from the beginning of the solution where the entire expression: $-y$ is under the power of 3 because it's inside parentheses that are raised to the power of 3, meaning:

$(-y)^3$.

First we will perform the multiplication of fractions using the rule for multiplying fractions:

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

Let's apply this rule to the problem:

$3^x\cdot\frac{1}{3^{-x}}\cdot3^{2x}=\frac{3^x}{1}\cdot\frac{1}{3^{-x}}\cdot\frac{3^{2x}}{1}=\frac{3^x\cdot1\cdot3^{2x}}{1\cdot3^{-x}\cdot1}=\frac{3^x\cdot3^{2x}}{3^{-x}}$

where in the first stage we performed the multiplication of fractions and then simplified the resulting expression,

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

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

Let's apply this law to the numerator of the expression we got in the last stage:

$\frac{3^x\cdot3^{2x}}{3^{-x}}=\frac{3^{x+2x}}{3^{-x}}=\frac{3^{3x}}{3^{-x}}$

Now let's recall the law of exponents for division between terms with identical bases:

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

Let's apply this law to the expression we got in the last stage:

$\frac{3^{3x}}{3^{-x}}=3^{3x-(-x)}=3^{3x+x}=3^{4x}$

When we applied the above law of exponents carefully, this is because the term in the denominator has a negative exponent so we used parentheses,

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

$3^x\cdot\frac{1}{3^{-x}}\cdot3^{2x}=\frac{3^x\cdot3^{2x}}{3^{-x}} = \frac{3^{3x}}{3^{-x}}=3^{4x}$

Now let's recall the law of exponents for power to a power but in the opposite direction:

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

Let's apply this law to the expression we got in the last stage:

$3^{4x}=3^{4\cdot x}=\big(3^4\big)^x$

When we applied the above law of exponents instead of opening the parentheses and performing the multiplication between the exponents in the exponent (which is the direct way of the above law of exponents), we represented the expression in question as a term with an exponent in parentheses to which an exponent applies.

Therefore the correct answer is answer B.

We'll use the law of exponents for negative exponents:

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

Let's apply this law to the problem:

$5^4-(\frac{1}{5})^{-3}\cdot5^{-2}= 5^4-(5^{-1})^{-3}\cdot5^{-2}$

When we apply the above law of exponents to the second term from the left,

Next, we'll recall the law of exponents for power of a power:

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

Let's apply this law to the expression we got in the last step:

$5^4-(5^{-1})^{-3}\cdot5^{-2} = 5^4-5^{(-1)\cdot (-3)}\cdot5^{-2} = 5^4-5^{3}\cdot5^{-2}$

When we apply the above law of exponents to the second term from the left and then simplify the resulting expression,

Let's continue and recall the law of exponents for multiplication of terms with the same base:

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

Let's apply this law to the expression we got in the last step:

$5^4-5^{3}\cdot5^{-2} =5^4-5^{3+(-2)}=5^4-5^{3-2}=5^4-5^{1} =5^4-5$

When we apply the above law of exponents to the second term from the left and then simplify the resulting expression,

From here we can notice that we can factor the expression by taking out the common factor 5 from the parentheses:

$5^4-5 =5(5^3-1)$

When we also used the law of exponents for multiplication of terms with the same base mentioned earlier, but in the opposite direction:

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

To notice that:

$5^4=5\cdot 5^3$

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

$5^4-(\frac{1}{5})^{-3}\cdot5^{-2}= 5^4-(5^{-1})^{-3}\cdot5^{-2} = 5^4-5^{3}\cdot5^{-2}=5(5^3-1)$

Therefore the correct answer is answer C.

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.

Let's handle each term in the initial expression separately:

a. We'll start with the leftmost term, meaning the exponent on the multiplication in parentheses,

We'll use the power rule for exponents on multiplication in parentheses:

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

which states that when an exponent applies to a multiplication in parentheses, it applies to each term in the multiplication when opening the parentheses,

Let's apply this to our problem for the leftmost term:

$(g\cdot a\cdot x)^4=g^4\cdot a^4\cdot x^4=g^4a^4x^4$

where in the final step we dropped the multiplication sign and switched to the conventional multiplication notation by placing the terms next to each other.

We're done with the leftmost term, let's move on to the next term.

b. Let's continue with the second term from the left, using the power rule for exponents:

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

Let's apply this rule to the second term from the left:

$(4^a)^x=4^{ax}$

and we're done with this term as well,

Let's summarize the results from a and b for the two terms in the initial expression:

$(g\cdot a\cdot x)^4+(4^a)^x=g^4a^4x^4+4^{ax}$

Therefore, the correct answer is c.

Notes:

a. For clarity and better explanation, in the solution above we handled each term separately. However, to develop proficiency and mastery in applying exponent rules, it is recommended to solve the problem as one unit from start to finish, where the separate treatment mentioned above can be done in the margin (or on a separate draft) if unsure about handling a specific term.

b. From the stated power rule for parentheses mentioned in solution a, it might seem that it only applies to two terms in parentheses, but in fact, it is valid for any number of terms in a multiplication within parentheses, as demonstrated in this problem and others,

It would be a good exercise to prove that if this rule is valid for exponents on multiplication of two terms in parentheses (as stated above), then it is also valid for exponents on multiplication of multiple terms in parentheses (for example - three terms, etc.).

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.

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.

In solving the problem, we use two laws of exponents, which we will mention:

a. The law of exponents for multiplying powers with the same bases:

$a^m\cdot a^n=a^{m+n}$b. The law of exponents for a power of a power:

$(a^m)^n=a^{m\cdot n}$We will apply these two laws of exponents in solving the problem in two steps:

Let's start by applying the law of exponents mentioned in a' to the first expression on the left side of the problem:

$10^{-3}\cdot10^4=10^{-3+4}=10^1=10$When in the first step we applied the law of exponents mentioned in a' and in the following steps we simplified the expression that was obtained,

We continue to the next step and apply the law of exponents mentioned in b' and handle the third expression on the left side of the problem:

$(4^2)^5=4^{2\cdot5}=4^{10}$When in the first step we applied the law of exponents mentioned in b' and in the following steps we simplified the expression that was obtained,

We combine the two steps detailed above to the complete problem solution:

$10^{-3}\cdot10^4-(7\cdot9\cdot5)^3+(4^2)^5= 10-(7\cdot9\cdot5)^3+4^{10}$In the next step we calculate the result of multiplying the numbers inside the parentheses in the second expression on the left:

$10-(7\cdot9\cdot5)^3+4^{10}= 10-315^3+4^{10}$Therefore, the correct answer is answer b'.

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.