Question

How do you solve `ln(x - 3) + ln(x + 4) = 1`?

Answer 1

`x~=3.37`

Explanation:

First thing to note is that a `ln(x-3)` is defined only when `x-3>0=>x>3`

And also `ln(x+4)` is defined when `x+4>0 => x> -4`

And so for both functions to be defined we need to intersect their domains, leading to : `x>3`

Now, that we have the domain we can solve the equation in this domain, thus:

`ln(x-3)+ln(x+4)=1`

We apply the logarithm laws `=> ln[(x-3)(x+4)]=1`

`=>(x-3)(x+4)=e`

`e` is Euler's constant

`=>x^2+x-12=e`

`=>x^2+x-12-e=0`

We apply the quadratic formula,

`=>x=(-1+-sqrt(1-4(-12-e)))/2=(-1+-sqrt(49+4e))/2`

The solution `(-1-sqrt(49+4e))/2` has to be rejected because it is negative and so falls out our domain `x>3`

The only solution is `x=(-1+sqrt(49+4e))/2~=color(blue)3.37`