Question

What is the value of ${{e}^{1.8}}$?

Answer 1

Hint: First of all we will understand the meaning of the term e. Now, consider the value of e is 2.718 and write the given expression ${{e}^{1.8}}={{e}^{2}}\times {{e}^{-0.2}}$. Calculate the value of ${{e}^{2}}$ by simply substituting the value of e. To calculate the value of ${{e}^{-0.2}}$ first write the expansion formula of ${{e}^{x}}$ given by the Maclaurin series as ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{5}}}{5!}+....$. Substitute the value of x equal to -0.2 and calculate the value up to 4 or 5 terms before neglecting the above terms. Finally, take the product of the value of ${{e}^{2}}$ and ${{e}^{-0.2}}$ to get the answer.

Complete step-by-step solution:
Here we have been asked to find the value of the expression ${{e}^{1.8}}$. First we need to understand the meaning of e.
Now, in mathematics e is called the Euler’s number and is a mathematical constant approximately equal to 2.718. e is also the base of the natural logarithm. As a sum of infinite series e can be given as $e=1+\dfrac{1}{1}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{5!}+....$ up to infinity. This expansion series is obtained by using the Maclaurin expansion formula. In general for any exponent (x)of e the expansion formula is ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{5}}}{5!}+....$.
Let us come to the question. We have the expression ${{e}^{1.8}}$ which can be written as ${{e}^{1.8}}={{e}^{2}}\times {{e}^{-0.2}}$. Calculating the value of ${{e}^{2}}$ we get,
$\begin{align}
  & \Rightarrow {{e}^{2}}={{\left( 2.718 \right)}^{2}} \\
 & \Rightarrow {{e}^{2}}\approx 7.38752 \\
\end{align}$
Now, we have to calculate the value of ${{e}^{-0.2}}$, so using the Maclaurin series and substituting the value of x = -0.2 we get,
$\begin{align}
  & \Rightarrow {{e}^{-0.2}}=1+\left( -0.2 \right)+\dfrac{{{\left( -0.2 \right)}^{2}}}{2!}+\dfrac{{{\left( -0.2 \right)}^{3}}}{3!}+\dfrac{{{\left( -0.2 \right)}^{4}}}{4!}+\dfrac{{{\left( -0.2 \right)}^{5}}}{5!}+.... \\
 & \Rightarrow {{e}^{-0.2}}=1+\left( -0.2 \right)+0.02+\left( -0.001\overline{3} \right)+\left( 0.0000\overline{6} \right)+\left( -0.000002\overline{6} \right)+.... \\
\end{align}$
The higher terms will be much smaller in value and can be neglected, so the above relation gives:
$\Rightarrow {{e}^{-0.2}}\approx 0.81873$
Multiplying the expressions ${{e}^{2}}$ and ${{e}^{-0.2}}$ we get,
\[\begin{align}
  & \Rightarrow {{e}^{2}}\times {{e}^{-0.2}}=7.38752\times 0.81873 \\
 & \Rightarrow {{e}^{.1.8}}\approx 6.04838 \\
\end{align}\]
Hence, the above obtained value is our approximate answer.

Note: Note that the value we have obtained is not exact but approximate. We cannot directly substitute x = 1.8 in the expansion formula of ${{e}^{x}}$ because 1.8 is a large number so we have to calculate many terms before we will start neglecting the higher terms. The other method to solve this question can be by taking the help of a log and antilog table. What we will do is we will assume ${{e}^{1.8}}=y$ and then take natural log both the sides which will give $\ln y=1.8$. Now, this relation can be converted into the common log given as $2.303{{\log }_{10}}y=1.8$. From here we will determine ${{\log }_{10}}y$ and then take antilog on both sides. Finally, we have to take the help of the antilog table to get the answer.