How do you solve for x in #log (x+6)=1-log(x-5)#?

Question

↳Redirected from "Question #e8623"

1996 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-09T13:18:28

#x_1=-(1+sqrt(161))/2#
#x_2=(-1+sqrt(161))/2#

#log(x+6)=1-log(x-5)#
then
#log(x+6)+log(x-5)=1#

using the rule of logaritm product

#log((x+6)(x-5))=1#

Remembering that Field of Existence of #log# is

#(x+6)(x-5)>0#

we find:

FE: # ]-oo,-6[ uu ]5,+oo[#

Now:

#10^(log((x+6)(x-5)))=10^1#

#(x+6)(x-5)=10#

#x^2-5x+6x-30=10#

#x^2+x-40=0#

#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

#x_(1,2)=(-1+-sqrt(1+160))/(2)=(-1+-sqrt(161))/(2)#

#x_1=-(1+sqrt(161))/2#
#x_2=(-1+sqrt(161))/2#

#x_(1,2) in ]-oo,-6[ uu ]5,+oo[#