Question #0be98

2 Answers
Nov 8, 2017

#x~~3.53 or x~~-4.53#

Explanation:

#log(x+3)+log(x-2)=1#

Using the addition rule, we get:
#log(x^2+x-6)=1#

#10^(log(x^2+x-6))=10^1=10#

#x^2+x-6=10#

#x^2+x-16=0#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-1-sqrt(1^2-4(-16)))/2=(-1-sqrt(65))/2~~3.53#

#x=(-1+sqrt(1^2-4(-16)))/2=(-1+sqrt(65))/2~~-4.53#

Nov 18, 2017

#x=-1/2+sqrt65/2~~3.53#

Explanation:

#log(x+3)+log(x-2)= 1#
#log[(x + 3)(x-2)]=1#
#x^2+x-6=10#
#x^2+x+(1/2)^2=10+6+1/4#
#(x+1/2)^2= 65/4#
#x+1/2=+-sqrt65/2#
#x=-1/2+-sqrt65/2#
#x~~ -4.53, 3.53# => reject the negative root, since it's not in the domain and makes the equation undefined thus it's an "Extraneous"
root.
therefore:
#x~~3.53# => the only valid solution