First thing to note is that a ln(x-3)ln(x−3) is defined only when x-3>0=>x>3x−3>0⇒x>3
And also ln(x+4)ln(x+4) is defined when x+4>0 => x> -4x+4>0⇒x>−4
And so for both functions to be defined we need to intersect their domains, leading to : x>3x>3
Now, that we have the domain we can solve the equation in this domain, thus:
ln(x-3)+ln(x+4)=1ln(x−3)+ln(x+4)=1
We apply the logarithm laws => ln[(x-3)(x+4)]=1⇒ln[(x−3)(x+4)]=1
=>(x-3)(x+4)=e⇒(x−3)(x+4)=e
ee is Euler's constant
=>x^2+x-12=e⇒x2+x−12=e
=>x^2+x-12-e=0⇒x2+x−12−e=0
We apply the quadratic formula,
=>x=(-1+-sqrt(1-4(-12-e)))/2=(-1+-sqrt(49+4e))/2⇒x=−1±√1−4(−12−e)2=−1±√49+4e2
The solution (-1-sqrt(49+4e))/2−1−√49+4e2 has to be rejected because it is negative and so falls out our domain x>3x>3
The only solution is x=(-1+sqrt(49+4e))/2~=color(blue)3.37x=−1+√49+4e2≅3.37
