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

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Answer 1 · 2015-07-14T21:57:59

#x~=3.37#

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#

#=>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#