Applying Combined Exponents Rules: Binomial

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.

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

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:

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

$a^530^5y^5$

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

We use the formula:

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

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

We simplify:

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

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

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.

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 note that the first term in the multiplication in the problem has an exponent of 0, and any number (different from zero) raised to the power of zero equals 1, meaning:

$X^0=1$Therefore, we get that the expression in the problem is:

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= 1\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}=(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}$

Later, we will use the law of exponents for negative exponents:

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

And before we proceed to solve the problem let's understand this law in a slightly different, indirect way:

Let's note that if we treat this law as an equation (which it indeed is in every way), and multiply both sides of the equation by the common denominator which is:

$a^n$we get:

$a^{-n}=\frac{1}{a^n}\\ \frac{a^n}{1} =\frac{1}{a^n}\hspace{8pt} \text{/}\cdot a^n\\ a^n\cdot a^{-n}=1$

Where in the first stage we remembered that any number can be represented as itself divided by 1, we applied this to the left side of the equation, then we multiplied by the common denominator and to know by how much we multiplied each numerator (after reduction with the common denominator) we addressed the question "by how much did we multiply the current denominator to get the common denominator?".

Let's look at the result we got:

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

Meaning that $a^n,\hspace{4pt}a^{-n}$are reciprocal numbers, or in other words:

$a^n$ is reciprocal to $a^{-n}$ (and vice versa),

and specifically:

$a,\hspace{4pt}a^{-1}$ are reciprocal to each other,

We can apply this understanding to the problem if we also remember that the reciprocal of a fraction is obtained by switching the numerator and denominator, meaning that the fractions:

$\frac{a}{b},\hspace{4pt}\frac{b}{a}$ are reciprocal fractions - which can be easily understood, since their multiplication will clearly give the result 1,

And if we combine this with the previous understanding, we can easily conclude that:

$\big(\frac{a}{b}\big)^{-1}=\frac{b}{a}$

In other words, raising a fraction to the power of negative one will give a result that is the reciprocal fraction, obtained by switching the numerator and denominator.

Let's return to the problem, the expression we got in the last stage is:

$(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}$

We'll use what was explained earlier and note that the fraction in parentheses in the second term of the multiplication is the reciprocal fraction to the fraction in parentheses in the first term of the multiplication, meaning that:$\frac{13}{2}= \big(\frac{2}{13} \big)^{-1}$Therefore we can simply calculate the expression we got in the last stage by converting to a common base using the above understanding:

$(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} = (\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5}$

Where we actually just replaced the fraction in parentheses in the second term of the multiplication with its reciprocal raised to the power of negative one as mentioned earlier,

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

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

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

$(\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{(-1)\cdot(-5)}=(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}$

Where in the first stage we applied the above law of exponents and then simplified the expression that resulted,

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

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= (\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} = (\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}$

In the next stage we'll recall the law of exponents for multiplication of terms with identical bases:

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

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

$(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5} =(\frac{2}{13}\big)^{-2+5}=(\frac{2}{13}\big)^{3}$

Where in the first stage we applied the above law of exponents and then simplified the expression,

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

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= (\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}=(\frac{2}{13}\big)^{3}$

Therefore the correct answer is answer A.

Recall the law of exponents for multiplication between terms with identical bases:

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

We'll use this to deal with the fraction's denominator in the problem:

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

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

Remember that raising any number to the power of 0 gives the result 1, or mathematically:

$X^0=1$

Therefore the denominator that we obtain in the last stage is 1.

This means that:

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

Recall the law of exponents for an exponent of a product inside of parentheses is as follows:

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

Apply this law to the first term in the product:

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

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

Let's summarize everything we did:

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

Therefore the correct answer is answer C.