lnx=3-ln(x+1) Solve for x.

Question

anonymous · Answer

ln(x^2-x)=3
x^2-x=e^3
I'm stuck here.

callisto · Answer

Something is wrong with the sign?

anonymous · Answer

ln(x^2+x)=3
x^2+x=e^3
x(x+1)=e^3
x=e^3 or e^3-1

anonymous · Answer

but the answer is 4.01

callisto · Answer

No...
From x^2+x=e^3
The next step is to subtract e^3 from both sides, so you'll get
x^2 + x - e^3 =0
Now, use the quadratic formula to solve x.

anonymous · Answer

why i'm wrong?

anonymous · Answer

i understand your method. i get 4.01 or 41.2 (rej)

callisto · Answer

For (x+a)(x+b) = c
You can only solve it by putting it as (x+a) = c and (x+b) = c when c=0. The logic is that
if you have (x+a)(x+b) = 0, either one of the factor, or both factors, is 0, so you solve it by putting (x+a)=0 or (x+b)=0.

anonymous · Answer

oh right! thanks a lot! i think i get it.

callisto · Answer

You're welcome. By the way, may I correct my mistakes? It should be "if you have (x+a)(x+b) = 0, either one of the factor is 0, so you solve it by putting (x+a)=0 or (x+b)=0."