How do you solve #ln(x-3)+ln(x+4)=3ln2#?

2 Answers
Aug 20, 2015

I found #x=4#

Explanation:

You can first use the property of logs:
#lnx+lny=ln(x*y)#
so:
#ln(x-3)+ln(x+4)=ln((x-3)(x+4))#
then the other property:
#alnx=lnx^a#
so you get:
#ln((x-3)(x+4))=ln2^3#
Now take the exponential of both sides:
#e^(ln((x-3)(x+4)))=e^(ln2^3)#
cancelling #e# and #ln# you get:
#(x-3)(x+4)=2^3#
#x^2+4x-3x-12=8#
#x^2+x-20=0#
Using the Quadratic Formula:
#x_(1,2)=(-1+-sqrt(1+80))/2#
#x_(1,2)=(-1+-9)/2#
Two solutions:
#x_1=-5#
#x_2=4#
Try the two into the original equation and you'll find that the negative does not work.

Jan 12, 2016

Minutely different way of talking this problem

Explanation:

Addition of logs is the result of multiplication of the source numbers

So #color(brown)(ln(x-3)+ln(x+4))color(blue)( -> ln(color(white)(...)(x+3)(x+4)color(white)(...)))#

Multiplication of a log is the result of the source number being raised to a power

So #color(brown)(3ln(2))color(blue)(-> ln(2^3))#

Putting it all together we have

#color(brown)(ln(color(blue)((x+3)(x+4)))=ln(color(blue)(2^3)) #

#color(green)("As both side are direct logs of each other we can simply forget")##color(green)("about the logs and write:")#

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

Now it can be solved as in Gio's solution