Question

How do you solve $\ln x + \ln (x - 2) = 1$?

Answer 1

Hint: In this question we will use the properties of logarithm and apply it to both sides of the equation to simplify the equation and solve it for the value of $x$.

Formula used:
$({x_1},{x_2}) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2ac}}$, Where $(x,y)$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.

Complete step-by-step answer:
We have the question given us to as:
$ \Rightarrow \ln x + \ln (x - 2) = 1$
Now we know the property of logarithm that $\ln a + \ln b = \ln ab$ therefore, on using this property, we get:
$ \Rightarrow \ln [x(x - 2)] = 1$
Now on taking the antilog on both the sides, we get:
$ \Rightarrow x(x - 2) = {e^1}$
Now on multiplying the terms in the left-hand side, we get:
$ \Rightarrow x \times x - x \times 2 = e$
On simplifying we get:
$ \Rightarrow {x^2} - 2x = e$
On transferring the term $e$ from the right-hand side to the left-hand side, we get:
$ \Rightarrow {x^2} - 2x - e = 0$
Now the given equation is in the form of a quadratic equation therefore, to find the value of $x$, we will factor the equation.
Since the middle term cannot be directly split to factorize the equation, we will use the quadratic formula to get the solution.
Now we have $a = 1$, $b = - 2$and $c = - e$
The roots can be found out as:
$ \Rightarrow ({x_1},{x_2}) = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - e)} }}{{2(1)( - e)}}$
On simplifying, we get:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{2 \pm \sqrt {4 + 4e} }}{2}\]
On taking out $4$as common in the root, we get:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{2 \pm \sqrt {4(1 + e)} }}{2}\]
On taking the square root of $4$, we get:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{2 \pm 2\sqrt {1 + e} }}{2}\]
Now on taking out $2$as common from the numerator we get:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{2(1 \pm \sqrt {1 + e} )}}{2}\]
On simplifying, we get:
\[ \Rightarrow ({x_1},{x_2}) = 1 \pm \sqrt {1 + e} \]
Therefore, the roots are: \[{x_1} = 1 + \sqrt {1 + e} \] and \[{x_2} = 1 - \sqrt {1 + e} \]

Now since the logarithm of a negative number does not exist, we discard ${x_2}$ therefore, the root is:
\[{x_1} = 1 + \sqrt {1 + e} \], which is the required solution.

Note:
It is to be noted that the logarithm we are using has the base $10$, 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.