Applying Combined Exponents Rules: Binomial

To solve this problem, we'll simplify the expression $2^{10} \times 3^6 \times 2^5 \times 3^2$ using the rules of exponents. Here are the steps:

• Step 1: Apply the product of powers property to the base 2 terms. The expression $2^{10} \times 2^5$ simplifies to:

  $2^{10+5} = 2^{15}$

• Step 2: Apply the product of powers property to the base 3 terms. The expression $3^6 \times 3^2$ simplifies to:

  $3^{6+2} = 3^8$

• Step 3: Combine the simplified terms to form the complete simplified expression:

  $2^{15} \times 3^8$

Therefore, the simplified form of the equation is $2^{15} \times 3^8$.

To solve this problem, we'll follow these steps:

• Step 1: Identify and group the terms with the same base.

• Step 2: Apply the laws of exponents to simplify by adding the exponents of each base.

• Step 3: Write the simplified form.

Let's work through each step:

Step 1: We are given that $4^7 \times 5^3 \times 4^2 \times 5^4$.

Step 2: First, group the terms with the same base:

$4^7 \times 4^2$ and $5^3 \times 5^4$.

Step 3: Use the law of exponents, which states $a^m \times a^n = a^{m+n}$.

For the base 4: $4^7 \times 4^2 = 4^{7+2} = 4^9$.

For the base 5: $5^3 \times 5^4 = 5^{3+4} = 5^7$.

Therefore, the simplified form of the expression is $4^9 \times 5^7$.

To solve this problem, we'll apply the laws of exponents to simplify the expression $7^5 \times 2^3 \times 7^2 \times 2^4$.

Let's follow these steps:

• Step 1: Identify like bases.
  We have two like bases in the expression: 7 and 2.

• Step 2: Apply the product of powers rule for each base separately.
  For the base 7: $7^5 \times 7^2 = 7^{5+2} = 7^7$.
  For the base 2: $2^3 \times 2^4 = 2^{3+4} = 2^7$.

• Step 3: Combine the results.
  The expression simplifies to $7^7 \times 2^7$.

The simplified form of the original expression is therefore $7^7 \times 2^7$.

Let's simplify the expression $5^3 \times 2^4 \times 5^2 \times 2^3$ using the rules for exponents. We'll apply the product of powers rule, which states that when multiplying like bases, you can add the exponents.

• Step 1: Focus on terms with the same base.
  Combine $5^3$ and $5^2$. Since both terms have the base $5$, we apply the rule $a^m \times a^n = a^{m+n}$: $5^3 \times 5^2 = 5^{3+2} = 5^5$

• Step 2: Combine $2^4$ and $2^3$. Similarly, for the base $2$: $2^4 \times 2^3 = 2^{4+3} = 2^7$

After simplification, the expression becomes:
$5^5 \times 2^7$

To simplify the given expression $4^2 \times 3^5 \times 4^3 \times 3^2$, we will follow these steps:

• Step 1: Identify and group similar bases.

• Step 2: Apply the rule for multiplying like bases.

• Step 3: Simplify the expression.

Now, let's go through each step thoroughly:

Step 1: Identify and group similar bases:
We see two distinct bases here: 4 and 3.

Step 2: Apply the rule for multiplying like bases:
For base 4: Combine $4^2$ and $4^3$, using the rule $a^m \times a^n = a^{m+n}$.

Add the exponents for base 4: $2 + 3 = 5$, thus, $4^2 \times 4^3 = 4^5$.

For base 3: Combine $3^5$ and $3^2$, still using the same exponent rule.

Add the exponents for base 3: $5 + 2 = 7$, resulting in $3^5 \times 3^2 = 3^7$.

Step 3: Simplify the expression:
The simplified expression is $4^5 \times 3^7$.

Therefore, the final simplified expression is $4^5 \times 3^7$.

To simplify the equation $6^4 \times 2^3 \times 6^2 \times 2^5$, we will make use of the rules of exponents, specifically the product of powers rule, which states that when multiplying two powers that have the same base, you can add their exponents.

Step 1: Identify and group the terms with the same base.
In the expression $6^4 \times 2^3 \times 6^2 \times 2^5$, group the powers of 6 together and the powers of 2 together:

• Powers of 6: $6^4 \times 6^2$

• Powers of 2: $2^3 \times 2^5$

Step 2: Apply the product of powers rule.
According to the product of powers rule, for any real number $a$, and integers $m$ and $n$, the expression $a^m \times a^n = a^{m+n}$.

Apply this rule to the powers of 6:
$6^4 \times 6^2 = 6^{4+2} = 6^6$.

Apply this rule to the powers of 2:
$2^3 \times 2^5 = 2^{3+5} = 2^8$.

Step 3: Write down the final expression.
Combining our results gives the simplified expression: $6^6 \times 2^8$.

Therefore, the solution to the problem is $6^6 \times 2^8$.

To solve this problem, we'll use the product of powers property which states $a^m \times a^n = a^{m+n}$.

• Step 1: Simplify the expression by grouping the like bases. The original expression is $7^3 \times 5^2 \times 7^4 \times 5^3$.

• Step 2: Combine the exponents for each base. For base 7: $7^3 \times 7^4 = 7^{3+4} = 7^7$. For base 5: $5^2 \times 5^3 = 5^{2+3} = 5^5$.

• Step 3: Write the simplified expression. After combining the exponents, the expression becomes $7^7 \times 5^5$.

Thus, the solution to the problem is $7^7 \times 5^5$.

To solve this problem, we'll employ the power of a power rule in exponents, which states that $(a^m)^n = a^{m \times n}$.

Let's apply this rule to each part of the expression:

• Step 1: Simplify $(3^2)^4$
  According to the power of a power rule, this becomes $3^{2 \times 4} = 3^8$.

• Step 2: Simplify $(5^3)^5$
  Similarly, apply the rule here to get $5^{3 \times 5} = 5^{15}$.

After simplifying both parts, we multiply the results:

$3^8 \times 5^{15}$

Thus, the reduced expression is $\boxed{3^8 \times 5^{15}}$.

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:

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

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

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

$15^3x^3$

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 formula:

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

Therefore, we obtain:

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

$a^530^5y^5$

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

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:

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

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

We simplify:

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

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

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:

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

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:

$8^3\times1=8^3$

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.