Question

The minimum value of x log x is equal to.
A. e
B. $\dfrac{1}{e}$
C. $-\dfrac{1}{e}$
D.$\dfrac{2}{e}$

Answer 1

Hint: We can use the product rule of differentiation, given by $\dfrac{d\left( u.v \right)}{dx}=\dfrac{vd\left( u \right)}{dx}+\dfrac{ud\left( v \right)}{dx}$ . Then we can find the minimum value for x log x. For finding minimum value, we can equate $\dfrac{dy}{dx}=0$ . We can also use some basic logarithm formula to make the solution short and simple. We can use the below formulas:
$\begin{align}
  & y=\log x \\
 & {{e}^{y}}=x \\
 & \log x-\log y=\log \left( \dfrac{x}{y} \right) \\
 & \dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x} \\
\end{align}$

Complete step-by-step answer:
It is given in the question that we have to find out the minimum value of x log x.
Let us assume that,
y = x log x …………………(i)

We know that the product rule of differentiation is given by;
$\dfrac{d\left( u.v \right)}{dx}=\dfrac{vd\left( u \right)}{dx}+\dfrac{ud\left( v \right)}{dx}$
We can use this basic differentiation on equation (i) with respect to x. We can also apply $\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}$ . So, we will get,
$\begin{align}
  & \dfrac{dy}{dx}=\log x\dfrac{d\left( x \right)}{dx}+x\dfrac{d\left( \log x \right)}{dx} \\
 & \dfrac{dy}{dx}=\log x.1+\dfrac{x}{x} \\
 & \dfrac{dy}{dx}=\log x+1...........\left( ii \right) \\
\end{align}$

We know that log e = 1 from basic logarithm. So, replacing 1 as $\log e$ in equation (ii), we get;
$\dfrac{dy}{dx}=\log x+\log e.........\left( iii \right)$
From basic logarithm formula, we have $\log a+\log b=\log \left( ab \right)$. Using this basic logarithmic formula in equation (iii), we get;
$\dfrac{dy}{dx}=\log \left( ex \right)$

Now to find the minimum value, we will equate $\dfrac{dy}{dx}$value to 0.
If $\dfrac{dy}{dx}=0$, then it is minimum.

Now form calculus we know that to know which point is maximum and which point is minimum we have to double derivative the function. If the sign f’’ (x) is positive, then it is a point of minimum and if the sign is negative then it is a point of maximum.
So, to find a minimum $\dfrac{dy}{dx}$ must be equal to 0.
$\begin{align}
  & \dfrac{dy}{dx}=\log ex=0 \\
 & or \\
 & \log ex=0 \\
\end{align}$

From basic logarithmic formula, if y = log a, then ${{e}^{y}}=a$. So, applying this in log ex = 0, we get;
$\Rightarrow ex={{e}^{0}}....................\left( iv \right)$
Also, we know that the value of ${{e}^{0}}=1.$ So, putting the value of ${{e}^{0}}=1$in equation (iv), we get;
$\begin{align}
  & \Rightarrow ex=1 \\
 & \Rightarrow x=\dfrac{1}{e} \\
\end{align}$

Now double differentiating equation (i), with respect to x, we get;
y = log ex
From the first differentiation, which we have already calculated, we have;
$\begin{align}
  & \dfrac{dy}{dx}=\log x+\log e \\
 & or \\
 & \dfrac{dy}{dx}=\log ex \\
\end{align}$

Again differentiating $\dfrac{dy}{dx}$ with respect to x, we get;
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{ex}\times e \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x}....................\left( v \right) \\
\end{align}$

Also, we have already calculated the value of $x=\dfrac{1}{e}$ .
Putting the value of x in equation (v), we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=e............\left( vi \right)$
So, from equation (vi) it is clear that e > 0. So, at $x=\dfrac{1}{e}$, the value of y is minimum. So, on putting the value of $x=\dfrac{1}{e}$ in $x\log x$ from this calculation of $x=\dfrac{1}{e}$ we get,
$x\log x=\dfrac{1}{e}\log \dfrac{1}{e}$

As, $\dfrac{1}{e}$ can be written as ${{e}^{-1}}$, we get,
$\begin{align}
  & =\dfrac{1}{e}\log {{e}^{-1}} \\
 & =\dfrac{-1}{e}\log e............\left( vii \right) \\
\end{align}$

We know that the value of log e = 1. So, on putting the value of log e = 1 in equation (vii), we get,
$\begin{align}
  & =\dfrac{-1}{e}\times 1 \\
 & =\dfrac{-1}{e} \\
\end{align}$

Thus, from this, we have the minimum value of $x\log x$ is equal to $\dfrac{-1}{e}$ , and option (C) is the correct answer.

Note: Make sure to learn all the logarithm formulas to solve this type of questions. Especially in this question using log (x) = y then ${{\left( e \right)}^{y}}=x$ is the most confusing part but without using this formula the solution may become too difficult to solve.