# How to express as a power of a rational number with a positive exponent?

Exponents and powers are used to show very large numbers or very small numbers in a simplified manner. For example, if we have to show 2 × 2 × 2 × 2 in a simple way, then we can write it as 2^{4}, where 2 is the base and 4 is the exponent. The whole expression 24 is said to be power.

Power is a value or an expression that represents repeated multiplication of the same number or factor. Number of times the base is multiplied to itself is the value of the exponent.Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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**For examples:**

3^{2} = 3 raised to power 2 = 3 × 3 = 9

4^{3} = 4 raised to power 3 = 4 × 4 × 4 = 64

An exponent of a number represents the number of times the number is multiplied by itself. For example- 2 is multiplied by itself for n times:

2 × 2 × 2 × 2 × …..n times = 2^{n}

The above expression, 2^{n}, is said as 2 raised to the power n. Therefore, exponents are also called power or sometimes indices.

**General Form of Exponents**

Exponent represents that how many times a number should be multiplied by itself to get the result. Thus any number ‘b’ raised to power ‘p’ can be expressed as:

b^{p}= {b × b × b × b × …. × b} p times

Here b is any number and p is a natural number.

- b
^{p}is also called the pth power of b. - ‘b’ is the base and ‘p’ is the exponent or index or power.
- ‘b’ is multiplied ‘p’ times, and thereby exponentiation is the shorthand method of repeated multiplication.

**Laws of Exponents**

Let ‘b’ is any number or integer (positive or negative) and ‘p1’, ‘p2’ are positive integers, denoting the power to the bases.

**Multiplication Law**

It states that the product of two exponents with the same base and different powers equals to base raised to the sum of the two powers or integers.

b^{p1}× b^{p2}= b^{(p1+p2)}

**Division Law**

It states that if two exponents having the same bases and different powers are divided, then the results will be base raised to the difference between both powers.

b^{p1}÷ b^{p2}= b^{p1}/ b^{p2 }= b^{(p1-p2)}

**Negative Exponent Law**

If the base has a negative power, then it can be converted into its reciprocal but with positive power or integer to the base.

b^{-p }= 1/b^{p}

**Basic Rules of Exponents**

There are certain basic rules defined for exponents in order to solve the exponential expressions along with the other mathematical operations, for example, if there are the product of two exponents, it can be simplified to make the calculation easier and is known as product rule, let’s look at some of the basic rules of exponents,

Product Rule ⇢ a

^{n}× a^{m}= a^{n + m}Quotient Rule ⇢ a

^{n }/ a^{m}= a^{n – m}Power Rule ⇢ (a

^{n})^{m}= a^{n × m}or^{m}√a^{n }= a^{n/m}Negative Exponent Rule ⇢ a

^{-m}= 1/a^{m}Zero Rule ⇢ a

^{0}= 1One Rule ⇢ a

^{1}= a

### How to express as a power of a rational number with a positive exponent?

**Solution: **

Rational numbers are of the form p/q, where p and q are integers and q ≠ 0. Because of the underlying structure of numbers, p/q form, most individuals find it difficult to distinguish between fractions and rational numbers.

When a rational number is divided, the output is in decimal form, which can be either ending or repeating. 3, 4, 5, and so on are some examples of rational numbers as they can be expressed in fraction form as 3/1, 4/1, and 5/1.

Now for the answer of the Question, consider the following example:

{(-3/5)^{2}× (-3/5)^{4}}^{3}[(-3/5)

^{2}× (-3/5)^{4}]^{3}= [(-3/5)

^{2+4}]^{3}= [(-3/5)

^{6}]^{3}= (-3/5)

^{6×3}= (-3/5)

^{18}

**Similar Question**

**Question 1: Solve (2 ^{2}) × (7^{2})**

**Solution:**

Here when bases are different and powers are same

So as per the product rule we can write as a

^{n}× b^{n}= (a × b)^{n.}So 2

^{2 }× 7^{2}= (2 × 7)

^{2}= 14

^{2}= 196

**Question 2: Find the value of 5 ^{-2} × 1/5^{2}.**

**Solution:**

Here we have 5

^{-2}× 1/5^{-2}We will use Negative Exponent Rule ⇢ a

^{-m}= 1/a^{m}So we can write above eq. as

5

^{-2}× 1/5^{2}= 5

^{-2}× 5^{-2}= 5

^{-2 + (-2)}= 5

^{-2 – 2}= 5

^{-4}= 1/5

^{4}

**Question 3: Simplify (5/8) ^{-7} × (8/5)^{-5}**

**Solution:**

(5/8)

^{-7}× (8/5)^{-5}= 1/(5/8)

^{7}× (8/5)^{-5}{Negative Exponent Rule ⇢ a^{-m}= 1/a^{m}}= (8/5)

^{7}× (8/5)^{-5}= (8/5)

^{7 + (-5) }{Product Rule ⇢ a^{n}× a^{m}= a^{n + m}}= (8/5)

^{2}= 64/25