Power of a Quotient Rule for Exponents: Solving the problem

When raising any number to the power of 0 it results in the value 1, mathematically:

$X^0=1$

Apply this to both the numerator and denominator of the fraction in the problem:

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

Note that -36 is a power of the number 6:

$36=6^2$

Apply this to the denominator to obtain expressions with identical bases in both the numerator and denominator:

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

Recall the power rule for power of a power in order to simplify the expression in the denominator:

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

Recall the power rule for division between terms with identical bases:

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

Apply these two rules to the expression that we obtained above:

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

In the first stage we applied the power of a power rule and proceeded to simplify the expression in the exponent of the denominator term. In the next stage we applied the second power rule - 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:

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

We'll apply it to the expression that we obtained:

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

Let's summarize the various steps of our solution:

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

Therefore the correct answer is A.

To solve this problem, follow these steps:

Step 1: Simplify the first fraction:

The first expression is $\frac{35x \cdot y^7}{7xy}$.

• Cancel the common factor of $7$: $35 \div 7 = 5$.

• This simplifies to $\frac{5x \cdot y^7}{x \cdot y}$.

• Cancel the common factor of $x$: $x/x$ cancels to $1$.

• Cancel part of the $y$ terms: $y^7/y = y^{7-1} = y^6$.

• The result is $5y^6$.

Step 2: Simplify the second fraction:

The second expression is $\frac{8x}{5y}$.

• No common factors in the numerator and denominator, so it remains $\frac{8x}{5y}$.

Step 3: Multiply these simplified results:

Now, multiply the results from Step 1 and Step 2: $5y^6 \cdot \frac{8x}{5y}$.

• The factor of $5$ in $5y^6$ and $\frac{8x}{5y}$ cancels: $5/5 = 1$.

• This results in $y^6 \cdot \frac{8x}{y}$.

• Cancel part of the $y$ terms: $y^6/y = y^{6-1} = y^5$.

Thus, the simplified expression is $8xy^5$.

Therefore, the solution to the problem is $\mathbf{8xy^5}$.

Let’s begin by multiplying the two fractions using the rule of fraction multiplication: multiply the numerators together and the denominators together, while keeping the fraction bar in place.

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

Let's apply this rule to our problem and perform the multiplication between the fractions:

$\frac{8x^7y^3}{20}\cdot\frac{4}{2x^5y^2}=\frac{8\cdot4\cdot x^7y^3}{20\cdot2\cdot x^5y^2}=\frac{32x^7y^3}{40x^5y^2}$

In the first step, we multiplied the fractions using the rule above and then simplified the numerator and denominator of the resulting fraction.

Next, we apply the same rule in reverse, rewriting the fraction as a product of separate fractions, each containing only numbers or terms with the same base:

$\frac{32x^7y^3}{40x^5y^2}=\frac{32}{40}\cdot\frac{x^7}{x^5}\cdot\frac{y^3}{y^2}$

We did this so we could continue and simplify the expression using the law of exponents for division between terms with identical bases:

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

Proceed to apply this law to the last expression that we obtained:

$\frac{32}{40}\cdot\frac{x^7}{x^5}\cdot\frac{y^3}{y^2}=\frac{4}{5}\cdot x^{7-5}\cdot y^{3-2}=\frac{4}{5}\cdot x^2\cdot y^1=\frac{4}{5}x^2y$

In this step, we not only applied the law of exponents but also simplified the numerical fraction by noticing that both the numerator and denominator are divisible by 8. After simplifying, we rewrote the expression in standard form, removing the multiplication dots and placing the terms side by side.

Finally, we can rewrite the result as a single fraction, recalling that multiplying a number by a fraction is equivalent to multiplying it by the numerator of that fraction:

$\frac{4}{5}x^2y=\frac{4x^2y}{5}$

Let's summarize the solution to the problem, we obtain the following:

$\frac{8x^7y^3}{20}\cdot\frac{4}{2x^5y^2} =\frac{32x^7y^3}{40x^5y^2} =\frac{4x^2y}{5}$

Therefore the correct answer is answer D.

Important note:

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

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

$\frac{8x^7y^3}{20}\cdot\frac{4}{2x^5y^2} =\frac{4x^2y}{5}$

(This means we could have skipped rewriting the fraction as a product of smaller fractions, and even the initial step of multiplying the fractions together. Instead, we could have simplified directly across numerators and denominators.)

However, it’s important to emphasize that this shortcut only works because every term in both numerators and denominators, as well as between the two fractions themselves, is connected by multiplication. This allows us to combine everything under a single fraction bar, simplify, and apply the laws of exponents directly. Not every problem will meet these conditions, so this method should be used with care.