How to Solve ln(x+2)-ln(x+1)=1 Using Log Laws

[ibysaiyan]

Homework Statement

solve
ln(x+2)-ln(x+1)=1

Homework Equations

log laws

The Attempt at a Solution

hi there, well i tried to solve it but got stuck pretty much at the start
=> ln(x+2)/ln(x+1)=1
=>ln(x+2)/ln(x+1)= e^1
multiply out brackets and rearrange them?
is this what i should do next
OH wait does it become =>(x+2)/(x+1)=e , as ln disappears due to relation of y=lnx : x=e^y?
Thanks in Advance.

 

[Unto]

well, Ln A - Ln B = Ln (A/B)

Then if Ln (A/B) = 1, you can anti log both sides, and solve for x

so A/B = e^1

for example

 

[ibysaiyan]

oh yea, thanks a lot i got it.. but ugh i got stuck on another equation this time =/
its ln(x+3)+ln(x-1)= 0
my attempt:
since its log a+log b= log ab

ln(x^2+2x-3)= o
x^+2x-3= e^0 which is 1.
is this correct?

 

[Unto]

Yes but you have not solved for x yet.

 

[Mark44]

So far, so good, but you're not done.
ln(x^2 + 2x - 3) = 0
x^2 + 2x -3 = 1
x^2 + 2x -4 = 0
Now solve the quadratic. Keep in mind that for your original log expressions to be defined, x > - 3 and x > 1, which means that x > 1. If you get a value of x such that x <= 1, you have to discard it.

 

[ibysaiyan]

So far, so good, but you're not done.
ln(x^2 + 2x - 3) = 0
x^2 + 2x -3 = 1
x^2 + 2x -4 = 0
Now solve the quadratic. Keep in mind that for your original log expressions to be defined, x > - 3 and x > 1, which means that x > 1. If you get a value of x such that x <= 1, you have to discard it.

thanks for your reply, after factorization the values i get are x=0,-4 .
i don't quite get it, now what to do with the original expression =?

 

[Mark44]

Your factorization is incorrect.

 

[ibysaiyan]

*facepalm* oh no =/ lol , sorry give me a sec
x= -2+$$\sqrt{}5$$,-2-$$\sqrt{}5$$

 

[Dick]

*facepalm* oh no =/ lol , sorry give me a sec
x= -2+$$\sqrt{}5$$,-2-$$\sqrt{}5$$

Still not quite right. Can you check that once more? And you should check the roots in your original equation. One or both of them may not be valid solutions.

 

[ibysaiyan]

ah this is embarrassing its basic factorization =/ ,oo i don't know what i am doing wrong , i tried both methods to factorize it, sorry its just that my mind is not with me --> 3.52 am.

 

[Dick]

Use the quadratic formula. I assume you were doing that. You just got a number wrong. And again, don't forget to try and plug the roots back into the original equation and check that they actually work. You can get false roots.