Question

Find the value of ${{e}^{5}}$.

Answer 1

Hint: We need to first explain the general expansion form of the expression of ${{e}^{x}}$. Then we can apply the expansion formula for $x=5$. We use the concept of factorials to find the decimal values of the individual terms and find the final solution of ${{e}^{5}}$.

Complete step by step answer:
We have to use the formula for an infinite expansion series of exponents to find the decimal value of the terms. We know that the infinite series of exponent gives ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{x}^{n}}}{n!}+.....\infty $.
We can understand that the base for the ${{e}^{x}}$ expansion has to be $e$ and the condition for $x\in \mathbb{R}$.

Now we want to find the value of ${{e}^{5}}$ and that is why we use the replacement of $x=5$.
Putting the value, we get ${{e}^{5}}=1+5+\dfrac{{{5}^{2}}}{2!}+\dfrac{{{5}^{3}}}{3!}+.....+\dfrac{{{5}^{n}}}{n!}+.....\infty $.
Now we use the concept of factorial. Here the term $n!$ defines the notion of multiplication of first n natural numbers.
This means $n!=1\times 2\times 3\times ....\times n$.
Simplifying the expression, we get ${{e}^{5}}=1+5+\dfrac{{{5}^{2}}}{2}+\dfrac{{{5}^{3}}}{6}+\dfrac{{{5}^{4}}}{24}+.....\infty =148.41$.

Therefore, we get the value of ${{e}^{5}}$ as $148.41$.

Note: We have to be careful about the base of the exponent. The formula can only be applied for the exponential base. In the case of a different base, we need to use logarithm to change the base to use the exponential expansion.