Question

How do you solve $\ln \left( x \right)=x-2$?

Answer 1

Hint: In this question we will first rearrange the terms in the expression to get it in the form of the expression $f\left( x \right)=x{{e}^{x}}$ so that we can use the lambert $W$function and find the value of $x$. We will use the lambert $W$ function which states that when $f\left( x \right)=x{{e}^{x}}$ we can take the inverse of the function and get the value of $x$ by the expression $W\left( x{{e}^{x}} \right)=x$ to get the required value of $x$.

Complete step-by-step answer:
We have the expression as:
$\Rightarrow \ln \left( x \right)=x-2$
Since it is in the form of a natural log, we will remove the log. We can write the expression as:
$\Rightarrow x={{e}^{x-2}}$
Now we know the property of exponent that ${{e}^{a+b}}={{e}^{a}}{{e}^{b}}$ therefore, we will use this property and write the expression as:
$\Rightarrow x={{e}^{x}}{{e}^{-2}}$
On transferring the term ${{e}^{x}}$ from the right-hand side to the left-hand side, we get:
$\Rightarrow \dfrac{x}{{{e}^{x}}}={{e}^{-2}}$
Now we know the property of exponents that $\dfrac{1}{{{a}^{b}}}={{a}^{-b}}$ therefore, on using the property, we get:
$\Rightarrow x{{e}^{-x}}={{e}^{-2}}$
On multiplying both the sides of the expression by $-1$, we get:
$\Rightarrow -x{{e}^{-x}}=-{{e}^{-2}}$
Now in the left-hand side we have got the expression in the form of $x{{e}^{x}}$ where $x=-x$ therefore, on using the lambert $W$ function, we can write:
$\Rightarrow -x={{W}_{n}}\left( -{{e}^{-2}} \right)$
On multiplying both the sides of the expression by $-1$, we get:
$\Rightarrow x=-{{W}_{n}}\left( -{{e}^{-2}} \right)$, which is the required solution.

Note: It is to be noted that the logarithm we are using has the base $e$, the base is the number to which the log value has to be raised to, to get the original term. This is also called the antilog of the number which is the logical reverse of taking a log.
The most commonly used bases in logarithm are $10$ and $e$ which has a value of approximate $2.713...$
Logarithm is used to simplify a mathematical expression, it converts multiplication to addition, division to subtraction and exponents to multiplication.