When we see a number that is not $0$ raised to zero, the result will be $1$.

Property formula:

$a^0=1$

This property is also concerning algebraic expressions.

When we see a number that is not $0$ raised to zero, the result will be $1$.

Property formula:

$a^0=1$

This property is also concerning algebraic expressions.

Question 1

\( 5^0= \)

Question 2

\( 112^0=\text{?} \)

Question 3

\( 5^4\cdot(\frac{1}{5})^4=\text{?} \)

Question 4

\( (\frac{7}{4})^?=1 \)

Question 5

\( \frac{9\cdot3}{8^0}=\text{?} \)

$5^0=$

We use the power property:

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

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

$1$

$112^0=\text{?}$

We use the zero exponent rule.

$X^0=1$We obtain

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

1

$5^4\cdot(\frac{1}{5})^4=\text{?}$

**This problem can be solved using the properties of powers for a negative power, power over a power, and the property of powers for the product between terms with identical bases, which is the natural way of solving,**

**But here we prefer to solve it in another way that is a bit faster:**

To this end, the power by power law is applied to the parentheses in which the terms are multiplied, **but in the opposite direction**:

$x^n\cdot y^n=(x\cdot y)^n$Since in the expression in the problem **there is a multiplication between two terms with identical powers**, this law can be used in its opposite sense, so we will apply this property to the problem:

$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4$**Since the multiplication in the given problem is between terms with the same power, we could apply this law in the opposite direction** and write the expression as the multiplication of the bases of the terms in parentheses to which the same power is applied.

We will continue and simplify the expression in parentheses, we will do it quickly if we notice that in parentheses **there is a multiplication between two opposite numbers, then their product will give the result: 1,** we will apply this understanding to the expression we arrived at in the last step:

$\big(5\cdot\frac{1}{5}\big)^4 = 1^4=1$When in the first step we apply the previous understanding, and then use the fact that **raising the number 1 to any power will always give the result: 1**, which means that:

$1^x=1$**Summarizing** the steps to solve the problem, we get that:

$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4 =1$__Therefore, the correct answer is option b.__

1

$(\frac{7}{4})^?=1$

We use the fact that raising any number (except zero) to the power of zero will yield the result 1:

$X^0=1$Therefore, it is clear that:

$(\frac{7}{4})^0=1$__Therefore, the correct answer is option C.__

0

$\frac{9\cdot3}{8^0}=\text{?}$

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:

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

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

$3^3$

Question 1

\( (\frac{7}{125})^0=\text{?} \)

Question 2

\( (300\cdot\frac{5}{3}\cdot\frac{2}{7})^0=\text{?} \)

Question 3

\( 7^4\cdot8^3\cdot(\frac{1}{7})^4=\text{?} \)

Question 4

\( 7^x\cdot7^{-x}=\text{?} \)

Question 5

\( 5^{-3}\cdot5^0\cdot5^2\cdot5^5= \)

$(\frac{7}{125})^0=\text{?}$

We use the zero exponent rule.

$X^0=1$We obtain:

$\big( \frac{7}{125}\big)^0=1$__Therefore, the correct answer is option B. __

1

$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0=\text{?}$

We use the fact that raising any number (except zero) to the power of zero will give the result 1:

$X^0=1$Let's examine the expression of the problem:

$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0$**The expression in parentheses is clearly not 0** (it can be calculated numerically and verified)

Therefore, the result of raising to the power of zero will give the result 1, that is:

$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0 =1$__Therefore, the correct answer is option A. __

1

$7^4\cdot8^3\cdot(\frac{1}{7})^4=\text{?}$

We use the formula:

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

We decompose the fraction in 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, so we obtain:

$8^3\times1=8^3$

$8^3$

$7^x\cdot7^{-x}=\text{?}$

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

$a^m\cdot a^n=a^{m+n}$We apply the property to the problem:

$7^x\cdot7^{-x}=7^{x+(-x)}=7^{x-x}=7^0$When in the first stage we apply the mentioned power property and in the following stages we simplify the expression obtained in the exponent,

Subsequently, we use the zero power property:

$X^0=1$We obtain:

$7^0=1$We summarize the solution to the problem, we obtained that:

$7^x\cdot7^{-x}=7^{x-x}=7^0 =1$__Therefore, the correct answer is option B. __

$1$

$5^{-3}\cdot5^0\cdot5^2\cdot5^5=$

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:

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

$5^{-3}\cdot5^0\cdot5^2\cdot5^5=5^{-3+0+2+5}=5^4$**Therefore, the correct answer is option c.**

**Note:**

Keep in mind that $5^0=1$

$5^4$

Question 1

\( 4^0=\text{?} \)

Question 2

\( (\frac{1}{8})^0=\text{?} \)

Question 3

\( 1^0= \)

Question 4

\( (0.1)^0= \)

Question 5

\( \frac{1}{5^0}= \)

$4^0=\text{?}$

$1$

$(\frac{1}{8})^0=\text{?}$

1

$1^0=$

$1$

$(0.1)^0=$

1

$\frac{1}{5^0}=$

1