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Question: How do you simplify \(3{{x}^{-3}}\) and write it using only positive exponents?...

How do you simplify 3x33{{x}^{-3}} and write it using only positive exponents?

Explanation

Solution

We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities to find the simplified form of 3x33{{x}^{-3}} with positive exponents.

Complete step by step solution:
We know the exponent form of the number aa with the exponent being nn can be expressed as an{{a}^{n}}. In case the value of nn becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is an=1an,nR+{{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}.
If we take two exponential expressions where the exponents are mm and nn.
Let the numbers be am{{a}^{m}} and an{{a}^{n}}. We take multiplication of these numbers.
The indices get added. So, am+n=am×an{{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}.
The division works in an almost similar way. The indices get subtracted. So,
aman=amn\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}.
We also have the identity of amn=(am)n{{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}.
For our given expression 3x33{{x}^{-3}}, we use the identity an=1an,nR+{{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}.
For given equation 3x33{{x}^{-3}}, we only take the part of x3{{x}^{-3}} as 3x3=3×x33{{x}^{-3}}=3\times {{x}^{- 3}}.
Using the identity, we get x3=1x3{{x}^{-3}}=\dfrac{1}{{{x}^{3}}}.
Multiplying with 3 we get 3x3=3x33{{x}^{-3}}=\dfrac{3}{{{x}^{3}}}.
Therefore, the simplified form of 3x33{{x}^{-3}} using only positive exponents is 3x3\dfrac{3}{{{x}^{3}}}.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers am{{a}^{m}} and an{{a}^{n}}, we have am±an=am(1±anm){{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right).the relation is independent of the values of mm and nn. We need to remember that the condition for am=anm=n{{a}^{m}}={{a}^{n}}\Rightarrow m=n is that the value of a0,±1a\ne 0,\pm 1.