# What is the domain of definition of the function f given by f(x)=square root (ln^2x-3lnx+2) ?

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### 2 Answers

We'll have to determine the values of x, for the square root and logarithm to make sense.

We'll start by imposing the condition for the radicand:

(ln x)^2 - 3ln x + 2 >= 0

When solving the inequality, we'll consider the following constraint of logarithm:

x>0

We'll solve the inequality, replacing ln x by t:

t^2 - 3t + 2 >= 0

We'll write the quadratic as a product of linear factors:

t^2 - 3t + 2 = (t - 1)(t - 2)

The values of t that makes the inequality positive are:

(0,1] U [2,+infinite)

For t = 1 => ln x = 1 => x = e

For t = 2 => ln x = 2 => x = e^2

**The domain of definition of the given function is represented by the sets: (0,e] U [e^2,+infinite).**

The domain of a function f(x) is the set of values of x for which the value of f(x) is real and defined.

For the function f(x)=`sqrt((ln^2x-3lnx+2))` , firstly (ln^2x-3lnx+2) cannot be negative as the square root of a negative number is a complex number.

This gives: `(ln x)^2-3*lnx+2 >= 0`

`(ln x)^2 - 2*ln x - ln x + 2 >= 0`

`ln x*(ln x - 2) - 1(ln x - 2) >= 0`

`(ln x - 1)(ln x - 2) >= 0`

For this to be true either both of the terms in the product should be positive or both of them should be negative.

ln x - 1 >= 0 and ln x - 2 >= 0

ln x >= 1 and ln x >= 2

x >= e^1 and x >= e^2

This is true for x >= e^2

ln x - 1 < 0 and ln x - 2 < 0

ln x < 1 and ln x <2

x < e and x < e^2

As the logarithm is defined only for positive numbers 0 < x < e

The domain of the given function is `(0, e)U[e^2, oo)`