Multiplication of Powers: Solving the problem

The problem requires simplification $\frac{9^2 \cdot 3^{-4}}{6^3}$ using exponent rules. Here’s a step-by-step guide to solving it:

• Step 1: Convert each term to powers of a common base.

  Notice that $9$ is $3^2$ and $6$ is $2 \times 3$. Hence:

  $9^2 = (3^2)^2 = 3^{4}$

  Therefore, the expression becomes $\frac{3^4 \cdot 3^{-4}}{(2 \cdot 3)^3}$.

• Step 2: Simplify the numerator.

  Using the exponent multiplication rule: $3^4 \cdot 3^{-4} = 3^{4 + (-4)} = 3^0 = 1$.

• Step 3: Expand the denominator.

  Calculate $(2 \cdot 3)^3$ by applying the distributive property: $(2 \cdot 3)^3 = 2^3 \cdot 3^3$.

• Step 4: Simplify the expression.

  After simplifying, the entire expression is $\frac{1}{2^3 \cdot 3^3}$.

  This simplifies further to $\frac{1}{6^3} = 6^{-3}$, because $2^3 \cdot 3^3 = (2 \cdot 3)^3 = 6^3$.

Therefore, the solution to the problem is $6^{-3}$.

Let's start by simplifying the second term in the complete multiplication, meaning - the fraction. We'll simplify it in two stages:

In the first stage we'll use the power law for multiplication between terms with identical bases:

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

and simplify the fraction's numerator:

$\frac{19^{35}\cdot19^{-32}}{19^3}=\frac{19^{35+(-32)}}{19^3}=\frac{19^{35-32}}{19^3}=\frac{19^3}{19^3}$

Next, we can either remember that dividing any number by itself gives 1, or use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ to get that:$\frac{19^3}{19^3}=19^{3-3}=19^0=1$

where in the last step we used the fact that raising any number to the power of 0 gives 1, meaning mathematically that:

$X^0=1$

Let's summarize this part, we got that:

$\frac{19^{35}\cdot19^{-32}}{19^3}=1$

Let's now return to the complete expression in the problem and substitute this result in place of the fraction:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}\cdot1=3^{-3}$

In the next stage we'll recall the power law for negative exponents:

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

and apply this law to the result we got:

$3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

Summarizing all the steps above, we got that:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}=\frac{1}{27}$

Therefore the correct answer is answer A.

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.

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

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

$3^2\times3^1=3^{2+1}=3^3$

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 $\mathbf{8xy^5}$.

Previously mentioned in the order of operations that exponents come before multiplication and division which come before addition and subtraction (and parentheses always come first),

Let's calculate therefore first the value of the expression inside the parentheses in the first term from the left (by calculating the values of the terms in the root inside the parentheses first) , let's handle the expression inside the parentheses separately first:

$\sqrt{8}\cdot\sqrt{2}-\sqrt{9}$Let's recall the definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$And we'll convert the roots in the first term to powers:

$\sqrt{8}\cdot\sqrt{2}-\sqrt{9} =8^{\frac{1}{2}}\cdot2^{\frac{1}{2}}-\sqrt{9}$

Next, we'll note that the two multiplication terms in the first term from the left have the same power, so let's recall the law of exponents for powers applied to parentheses, 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 we can write a multiplication of two terms with equal exponents as a multiplication of the bases in parentheses raised to that same power, let's apply this to the expression we got in the last step:

$8^{\frac{1}{2}}\cdot2^{\frac{1}{2}}-\sqrt{9} =(8\cdot2)^{\frac{1}{2}}-\sqrt{9}=16^{\frac{1}{2}}-\sqrt{9}$

Where in the first stage we applied the above law of exponents and in the second stage we calculated the result of the multiplication in parentheses,

Let's return to writing roots using the definition of root as a power, but in the opposite direction:

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

And we'll convert back the power of one-half to a square root (while remembering that a square root is actually a root of order 2), then we'll simplify the expression by calculating the values of the roots (without a calculator!):

$16^{\frac{1}{2}}-\sqrt{9} =\sqrt{16}-\sqrt{9}=4-3=1$

Where in the last stage we calculated the result of the subtraction operation,

Let's summarize then the development stages for the expression in parentheses in the original problem, we got that:

$\sqrt{8}\cdot\sqrt{2}-\sqrt{9} = (8\cdot2)^{\frac{1}{2}}-\sqrt{9}= \sqrt{16}-\sqrt{9}=4-3=1$

Let's return now to the expression in the original problem and substitute this result:

$(\sqrt{8}\cdot\sqrt{2}-\sqrt{9})^2:2^2+3^2 \\ \downarrow\\ 1^2:2^2+3^2$

We'll continue and after we calculate the values of the numbers in powers, we'll perform the division operation and then the addition operation:

$1^2:2^2+3^2=1:4+9=\frac{1}{4}+9=9\frac{1}{4}$

Where in the second stage we converted the division operation to fraction notation,

Therefore the correct answer is answer C.

First let's note that the number 9 is a power of the number 3:

$9=3^2$

Therefore we can immediately move to a unified base in the problem, in addition we'll recall the law of powers for negative exponents but in the opposite direction:

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

Let's apply this to the problem:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1}$

In the first term of the multiplication we replaced the number 9 with a power of 3, according to the relationship mentioned earlier, and simultaneously the third term in the multiplication we expressed as a term with a negative exponent according to the aforementioned law of exponents.

Now let's recall two additional laws of exponents:

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

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

b. The law of exponents for multiplication between terms with equal bases:

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

Let's apply these two laws to the expression we got in the last stage:

$(3^2)^4\cdot3^{-8}\cdot3^{-1}=3^{2\cdot4}\cdot3^{-8}\cdot3^{-1}=3^8\cdot3^{-8}\cdot3^{-1}=3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

In the first stage we applied the law of exponents for power of a power mentioned in a', in the next stage we applied the law of exponents for multiplication of terms with identical bases mentioned in b', then we simplified the resulting expression.

Let's summarize the solution steps:

$9^4\cdot3^{-8}\cdot\frac{1}{3}=(3^2)^4\cdot3^{-8}\cdot3^{-1} =3^{8+(-8)+(-1)}=3^{8-8-1}=3^{-1}$

Therefore the correct answer is answer b'.