
How do you simplify $ {{e}^{-2}}.\text{ }{{e}^{6}} $ ?
Answer
563.1k+ views
Hint: This question can be solved by using the law of exponents. Laws of exponents are as follows:
Product of powers: $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $
Quotient of powers: $ {{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} $
Power of a power: $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $
Power of a product: $ {{\left( xy \right)}^{m}}={{x}^{m}}\times {{y}^{m}} $
Power of a quotient: $ {{\left( \dfrac{x}{y} \right)}^{m}}=\dfrac{{{x}^{m}}}{{{y}^{m}}} $
Negative exponent: $ {{x}^{-m}}=\dfrac{1}{{{x}^{m}}} $ or $ {{x}^{m}}=\dfrac{1}{{{x}^{-m}}} $
Identity exponent: $ {{x}^{1}}=x $
Zero exponent: $ {{x}^{0}}=1(x\ne 0) $
We need to write the final answer in exponential form only because our base is an exponent (e) itself and not a number here.
Complete step by step answer:
Now, let’s solve the question.
First, write the expression given in the question:
$ \Rightarrow {{e}^{-2}}.\text{ }{{e}^{6}} $
As we can see here, the bases are the same and the powers are different and are in multiplication form. So we need to apply the product of powers rule in the above expression.
Product of power rule: $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $
After applying the rule:
$ \Rightarrow {{e}^{-2+6}} $
$ \therefore {{e}^{4}} $
So, this is our final answer. Leave the answer in the exponential form only as the base is not the number. It is a natural base.
Note:
There is an alternative way to solve this question. Let’s see that too. First, write the expression given in the question:
$ \Rightarrow {{e}^{-2}}.\text{ }{{e}^{6}} $
Now write the negative exponent or we can say multiplicative inverse of $ {{e}^{-2}} $ .
$ \Rightarrow \dfrac{1}{{{e}^{2}}}.\text{ }{{e}^{6}} $
We can write the above expression as:
$ \Rightarrow \dfrac{{{e}^{6}}}{{{e}^{2}}} $
As we can see here, the bases are the same and the powers are different and are in division form. So we need to apply the quotient of powers rule in the above expression.
Quotient of power rule is: $ {{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} $
After applying the rule:
$ \begin{align}
& \Rightarrow {{e}^{6-2}} \\
& \therefore {{e}^{4}} \\
\end{align} $
So, you can see that we have got the same answers by applying both the methods. Before solving such type of questions, you should know all the laws of exponents.
Product of powers: $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $
Quotient of powers: $ {{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} $
Power of a power: $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $
Power of a product: $ {{\left( xy \right)}^{m}}={{x}^{m}}\times {{y}^{m}} $
Power of a quotient: $ {{\left( \dfrac{x}{y} \right)}^{m}}=\dfrac{{{x}^{m}}}{{{y}^{m}}} $
Negative exponent: $ {{x}^{-m}}=\dfrac{1}{{{x}^{m}}} $ or $ {{x}^{m}}=\dfrac{1}{{{x}^{-m}}} $
Identity exponent: $ {{x}^{1}}=x $
Zero exponent: $ {{x}^{0}}=1(x\ne 0) $
We need to write the final answer in exponential form only because our base is an exponent (e) itself and not a number here.
Complete step by step answer:
Now, let’s solve the question.
First, write the expression given in the question:
$ \Rightarrow {{e}^{-2}}.\text{ }{{e}^{6}} $
As we can see here, the bases are the same and the powers are different and are in multiplication form. So we need to apply the product of powers rule in the above expression.
Product of power rule: $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $
After applying the rule:
$ \Rightarrow {{e}^{-2+6}} $
$ \therefore {{e}^{4}} $
So, this is our final answer. Leave the answer in the exponential form only as the base is not the number. It is a natural base.
Note:
There is an alternative way to solve this question. Let’s see that too. First, write the expression given in the question:
$ \Rightarrow {{e}^{-2}}.\text{ }{{e}^{6}} $
Now write the negative exponent or we can say multiplicative inverse of $ {{e}^{-2}} $ .
$ \Rightarrow \dfrac{1}{{{e}^{2}}}.\text{ }{{e}^{6}} $
We can write the above expression as:
$ \Rightarrow \dfrac{{{e}^{6}}}{{{e}^{2}}} $
As we can see here, the bases are the same and the powers are different and are in division form. So we need to apply the quotient of powers rule in the above expression.
Quotient of power rule is: $ {{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} $
After applying the rule:
$ \begin{align}
& \Rightarrow {{e}^{6-2}} \\
& \therefore {{e}^{4}} \\
\end{align} $
So, you can see that we have got the same answers by applying both the methods. Before solving such type of questions, you should know all the laws of exponents.
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