How do you solve #log(x)+log(x+3)=1#?

Question

23038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T19:05:25

#x=2#

taking#" "logx-=log_10x#

using the law of logs

#logA+logB=logAB#

we have

#logx+log(x+3)=1#

#logx(x+3)=1--(1)#

definition of logs

#log_ab=c=>a^c=b#

#(1)rarr10^1=x(x+3)#

#:.x^2+3x=10#

#x^2+3x-10=0#

factorising and solving

#(x+5)(x-2)=0#

#=>x=-5, " or "x=2#

#because logx" is undefined for "x<=0#

#x=-5 " is not a solution"#

#x=2#