Applying Combined Exponents Rules: Binomial

We use the formula:

$(a\times b)^n=a^nb^n$

$(5\times x\times3)^3=(15x)^3$

$(15x)^3=(15\times x)^3$

$15^3x^3$

We use the formula:

$(a\times b)^x=a^xb^x$

Therefore, we obtain:

$(a\times5\times6\times y)^5=(a\times30\times y)^5$

$a^530^5y^5$

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

Let us begin by using the law of exponents for a power that is applied to parentheses in which terms are multiplied:

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

We apply the rule to our problem:

$(x\cdot4\cdot3)^3= x^3\cdot4^3\cdot3^3$

When we apply the power to the product of the terms within parentheses, we apply the power to each term of the product separately and keep the product,

Therefore, the correct answer is option C.

We use the power law for multiplication within parentheses:

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

We apply it in the problem:

$(y\cdot7\cdot3)^4=y^4\cdot7^4\cdot3^4$

Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we can understand that it applies not only to two terms within parentheses, but also for multiple terms within parentheses.

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.

We use the formula:

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

We decompose the fraction inside of the parentheses:

$(\frac{1}{7})^4=\frac{1^4}{7^4}$

We obtain:

$7^4\times8^3\times\frac{1^4}{7^4}$

We simplify the powers: $7^4$

We obtain:

$8^3\times1^4$

Remember that the number 1 in any power is equal to 1, thus we obtain:

$8^3\times1=8^3$

We use the formula:

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

Therefore, we obtain:

$(\frac{3}{2})^4$

We use the formula:

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

Therefore, we obtain:

$\frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}$

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

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

and we'll use it to handle the fraction's denominator in the problem:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=\frac{(-3)^5\cdot8^4}{(-3)^{3+2+(-5)}}=\frac{(-3)^5\cdot8^4}{(-3)^0}$

where in the first stage we'll apply the above law to the denominator and then simplify the expression with the exponent in the denominator,

Now let's remember that raising any number to the power of 0 gives the result 1, or mathematically:

$X^0=1$

therefore the denominator we got in the last stage is 1,

meaning we got that:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=\frac{(-3)^5\cdot8^4}{(-3)^0}=\frac{(-3)^5\cdot8^4}{1}=(-3)^5\cdot8^4$

Now let's recall the law of exponents for an exponent of a product in parentheses:

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

and we'll apply this law to the first term in the product we got:

$(-3)^5\cdot8^4=(-1\cdot3)^5\cdot8^4 =(-1)^5\cdot3^5\cdot8^4=-1\cdot 3^5\cdot 8^4=-3^5\cdot8^4$

Note that the exponent applies separately to both the number 3 and its sign, which is the minus sign that is actually multiplication by $-1$

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

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=(-3)^5\cdot8^4 = -3^5\cdot8^4$

Therefore the correct answer is answer C.