# What is the domain of f in f(x)=ln(x^2+4x-5) ?

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`f(x)= ln(x^2+4x-5)`

To determine the domain, apply the property of logarithm which is its argument is always positive `( ln a, agt0)` .

So, set x^2+4x-5 greater than zero.

`x^2+4x-5gt0`

Then, factor left side.

`(x+5)(x-1)gt0`

And set each factor equal to zero and solve for x.

For the first factor,

`x+5=0`

`x+5-5=0-5`

`x=-5`

For the second factor,

`x - 1 = 0`

`x -1 + 1=0+1`

`x=1`

Next, determine the intervals in which it satisfy the inequality equation, using the values of x as the boundary of each interval.

For the first interval `(-oo,-5)` use the test value x=-6 and substitute it to the inequality equation.

`x^2+4x-5gt0`

`(-6)^2+4(-6)-5gt0`

`36-24-5gt0`

`7gt0` (True)

For the second interval `(-5,1)` , use the test value x=0.

`0^2+4(0)-5gt0`

`0+0-5gt0`

`-5gt0` (False)

And for the third interval `(1,+oo)` , use the test value x=2.

`2^2+4(2)-5gt0`

`4+8-5gt0`

`7gt0` (True)

**Hence, the domain of the given function is `(-oo,-5)uu(1,+oo)` .**

funtion log(t) is defined for `t>0`

in this case `t=x^2+4x-5` thus is enough to find for wich values of x `t>0`

First step is to findthe root of equation:

`Delta= 16-4(-5)=36>0`

thus has two differete real solutions

`x=(-4+-sqrt(36))/2=(-4+-6)/2=-2+-3`

so:

`x_1=1` `x_2 = -5`

Then we can write `x^2+4x-5=(x-1)(x+5)`

Now this product is greater than zero if `x<-5` and `x>1`

so for this value `t>0` , that is function is defined.

note that in the closed segment [-5;1] funtion isn't deifned. (Both the point -5 and 1 include bicouse for this value the argument t =0)