How do you solve #Log(x+2)+log(x-1)=4#?

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Answer 1 · 2016-11-22T08:03:10

#x~=99.51#

#log(x+2)+log(x-1)=4#

#hArrlog((x+2)(x-1))=4#

i.e. #(x+2)(x-1)=10^4=10000#

or #x^2+x-10002=0#

and using quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-1+-sqrt(1+40008))/2#

Now, we cannot use #-# sign as that makes #log(x+2)# or #log(x-1)# not possible.

Hence #x=(-1+sqrt(1+40008))/2~=(-1+200.02)/2=199.02/2=99.51#