How do you solve #ln(x+4)-ln(x+3)=lnx#?

Question

5227 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-23T23:20:48

#x=-1+sqrt(5)#

Dominion: #color(blue)(x+4>0 and x+3>0 and x>0 <=> x>0#

#ln(x+4)-ln(x+3)=ln(x)#

#ln(x+4)=ln(x+3)+ln(x)#

The sum of logarithms is the logarithm of the product

#ln(x+4)=ln((x+3)x)#

#color(blue)(ln(a)=ln(b) ->a=b#

#x+4=(x+3)x#

#x+4=x^2+3x #

#x^2+2x-4 =0#

#x=(-2+-sqrt(2^2-4*1*(-4)))/2#

#x=(-2+-sqrt(4+16))/2#

#x=(-2+-sqrt(20))/2#

#x=(-2+-2sqrt(5))/2#

#x=-1+-sqrt(5)#

#x=-1+sqrt(5)#, because x must be larger than 0