What is the value of \[{e^0}\] and \[{e^1}\]?
Answer
541.5k+ views
Hint: We will first understand what \[e\] stands for and then apply some basic exponential rules to get the solution. The \[e\] written here, in the question, is an Euler’s number or an Euler’s constant or an exponential constant. The value of \[e\] is an irrational number, which means the number goes to infinity after the decimal point.
Complete step-by-step solution:
You might have noticed this \[e\] as a base of log i.e., \[e\] acts as a base of logarithmic function and \[\log e = 1\]. We can say that \[{\left( {1 + \dfrac{1}{n}} \right)^n}\] is the limit of Euler’s number where n stands for an infinite number. Only when the value of n approaches infinity, then the value \[{\left( {1 + \dfrac{1}{n}} \right)^n}\] will reach e.
The approximate value of \[e\] is 2.718 but it does not end here. When the power reaches infinity then the value of \[e\] becomes zero. It can be expressed as below
\[e = \sum {\dfrac{1}{{n!}}} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{{1.2}} + \dfrac{1}{{1.2.3}} + \dfrac{1}{{1.2.3.4}} + ....\]
So, we have looked at the way of deriving the value of \[e\]. But we will not go so deep with this concept. We just have to remember the basic value of \[e\] that is 2.718. Now we will find the value of \[{e^0}\] and \[{e^1}\].
According to the exponential rule, we can say that \[{e^0}\] is similar to \[{a^0}\]. That means,
\[{e^0} = {a^0}\]
Since we already know that anything raised to the power 0 is 1. Therefore, we can say that
\[{a^0} = 1\] and hence,
\[{e^0} = 1\]
Now, we have to find the value of \[{e^1}\] and again by the rule of exponential we can assume that this
\[{e^1} = {a^1}\]
And we know that anything raised to the power 1 is itself and hence
\[{e^1} = e\]
Since we know the value of \[e\] is 2.718, therefore the value of \[{e^1}\] is
\[{e^1} = e = 2.718\]
Therefore, \[{e^0} = 1\] and \[{e^1} = 2.718\]
Note: We don’t need to memorize the full value of \[e\] and can remember the value to be 2.718. Having the knowledge about exponential rules is a must to solve these types of questions. In schools, we don’t need much knowledge about it, but this helps us to measure the non-linear increase and decrease of a function.
Complete step-by-step solution:
You might have noticed this \[e\] as a base of log i.e., \[e\] acts as a base of logarithmic function and \[\log e = 1\]. We can say that \[{\left( {1 + \dfrac{1}{n}} \right)^n}\] is the limit of Euler’s number where n stands for an infinite number. Only when the value of n approaches infinity, then the value \[{\left( {1 + \dfrac{1}{n}} \right)^n}\] will reach e.
The approximate value of \[e\] is 2.718 but it does not end here. When the power reaches infinity then the value of \[e\] becomes zero. It can be expressed as below
\[e = \sum {\dfrac{1}{{n!}}} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{{1.2}} + \dfrac{1}{{1.2.3}} + \dfrac{1}{{1.2.3.4}} + ....\]
So, we have looked at the way of deriving the value of \[e\]. But we will not go so deep with this concept. We just have to remember the basic value of \[e\] that is 2.718. Now we will find the value of \[{e^0}\] and \[{e^1}\].
According to the exponential rule, we can say that \[{e^0}\] is similar to \[{a^0}\]. That means,
\[{e^0} = {a^0}\]
Since we already know that anything raised to the power 0 is 1. Therefore, we can say that
\[{a^0} = 1\] and hence,
\[{e^0} = 1\]
Now, we have to find the value of \[{e^1}\] and again by the rule of exponential we can assume that this
\[{e^1} = {a^1}\]
And we know that anything raised to the power 1 is itself and hence
\[{e^1} = e\]
Since we know the value of \[e\] is 2.718, therefore the value of \[{e^1}\] is
\[{e^1} = e = 2.718\]
Therefore, \[{e^0} = 1\] and \[{e^1} = 2.718\]
Note: We don’t need to memorize the full value of \[e\] and can remember the value to be 2.718. Having the knowledge about exponential rules is a must to solve these types of questions. In schools, we don’t need much knowledge about it, but this helps us to measure the non-linear increase and decrease of a function.
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