How do you solve #Ln (x-1) + ln (x+2) = 1#?

Question

↳Redirected from "What did Rutherford's gold foil show about the structure of an atom?"

3181 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-20T19:17:38

#x = 1/2 (-1 +sqrt[9 + 4 e]) = 1.72896#

#Ln (x-1) + ln (x+2) = 1->Ln(x-1)(x+2)=Ln e#

#(x-1)(x+2)-e=0#

Solving for #x#

#x = 1/2 (-1 pm sqrt[9 + 4 e]) = {-2.72896, 1.72896}#

Now, substituting those values in #(x-1)# and #(x+2)# keeping in mind that both must be positive, we get at the solution.

#x = 1/2 (-1 +sqrt[9 + 4 e]) = 1.72896#