Exponents Rules - Examples, Exercises and Solutions

To solve this problem, we'll focus on expressing the power $2^7$ as a series of multiplications.

• Step 1: Identify the given power expression $2^7$.
• Step 2: Convert $2^7$ into a product of repeated multiplication. This involves writing 2 multiplied by itself for a total of 7 times.
• Step 3: The expanded form of $2^7$ is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$.

By comparing this expanded form with the provided choices, we see that the correct expression is:

$2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$

Therefore, the solution to the problem is the expression that matches this expanded multiplication form, which is the choice $\text{1: } 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$.

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Let's return to the problem:

First keep in mind that:

$10=10^1$Keep in mind that all the terms of the multiplication have the same base, so we will use the previous property:

$10^1\cdot10^2\cdot10^{-4}\cdot10^{10}=10^{1+2-4+10}=10^9$

$$Therefore, the correct answer is option c.

We use the zero exponent rule.

$X^0=1$We obtain

$112^0=1$Therefore, the correct answer is option C.

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

• Step 1: Set up the multiplication as $11 \times 11$.
• Step 2: Compute the product using basic arithmetic.
• Step 3: Compare the result with the provided multiple-choice answers to identify the correct one.

Now, let's work through each step:
Step 1: We begin with the calculation $11 \times 11$.
Step 2: Perform the multiplication:

1. Multiply the units digits: $1 \times 1 = 1$.
2. Next, for the tens digits: $11 \times 10 = 110$.
3. Add the results: $110 + 1 = 111$. This doesn't seem right, so let's break it down further.

Let's examine a more structured multiplication method:

Multiply $11$ by $1$ (last digit of the second 11), we get 11.
Multiply $11$ by $10$ (tens place of the second 11), we get 110.

If we align correctly and add the partial products:

     11
+   110
------------
   121

Step 3: The correct multiplication yields the final result as $121$. Upon reviewing the provided choices, the correct answer is choice 4: $121$.

Therefore, the solution to the problem is $121$.

In order to solve the exercise, we use the negative exponent rule.

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

We apply the rule to the given exercise:

$19^{-2}=\frac{1}{19^2}$

We can then continue and calculate the exponent.

$\frac{1}{19^2}=\frac{1}{361}$

We use the formula:

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

$(2^2)^3+(3^3)^4+(9^2)^6=2^{2\times3}+3^{3\times4}+9^{2\times6}=2^6+3^{12}+9^{12}$

To solve the given expression $(2^3)^6$, we apply the power of a power rule $(a^m)^n = a^{m \cdot n}$. Here, $a = 2$, $m = 3$, and $n = 6$.

Thus, we calculate the exponent:

$3 \cdot 6 = 18$

So, $(2^3)^6 = 2^{18}$.

To solve the problem, we need to apply the power of a product exponent rule. This rule states that when you raise a product to a power, it's the same as raising each factor to that power. In mathematical terms, if you have $(abc)^n$, it is equivalent to $a^n \times b^n \times c^n$.

Let's apply this rule step by step:

Our original expression is: $(2 \times 7 \times 5)^3$.

We first identify the factors inside the parentheses as $2$, $7$, and $5$.

According to the Power of a Product rule, we can distribute the exponent$3$ to each factor:

First, raise $2$ to the power of $3$ to get $2^3$.

Then, raise $7$ to the power of $3$ to get $7^3$.

Finally, raise $5$ to the power of $3$ to get $5^3$.

Therefore, the expression $(2 \times 7 \times 5)^3$ simplifies to $2^3 \times 7^3 \times 5^3$.

To solve the exercise we use the power property:$(a^n)^m=a^{n\cdot m}$

We use the property with our exercise and solve:

$(3^5)^4=3^{5\times4}=3^{20}$

We use the power law for multiplication within parentheses:

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

We apply it to the problem:

$(3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4$

Therefore, the correct answer is option b.

Note:

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

We begin by using the power rule of negative exponents.

$a^{-n}=\frac{1}{a^n}$We then apply it to the problem:

$$$4^{-1}=\frac{1}{4^1}=\frac{1}{4}$We can therefore deduce that the correct answer is option B.

We use the formula:

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

$(4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}$

To solve $(4^3)^2$, we use the power of a power rule which states that $(a^m)^n = a^{m \cdot n}$.

Here, $a = 4$, $m = 3$, and $n = 2$.

So, we calculate $4^{3 \cdot 2}$,

which simplifies to $4^6$.

We use the power property:

$X^0=1$We apply it to the problem:

$5^0=1$Therefore, the correct answer is C.

We use the property of powers of a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it to the problem:

$5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

Therefore, the correct answer is option d.