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

  • Thread starter ibysaiyan
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In summary, the conversation is about solving the equation ln(x+2)-ln(x+1)=1 using log laws and the quadratic formula. The solution involves factoring the equation and checking for valid roots.
  • #1
ibysaiyan
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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.
 
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  • #2
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
 
  • #3
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?
 
  • #4
Yes but you have not solved for x yet.
 
  • #5
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.
 
  • #6
Mark44 said:
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 =?
 
  • #8
*facepalm* oh no =/ lol , sorry give me a sec
x= -2+[tex]\sqrt{}5[/tex],-2-[tex]\sqrt{}5[/tex]
 
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  • #9
ibysaiyan said:
*facepalm* oh no =/ lol , sorry give me a sec
x= -2+[tex]\sqrt{}5[/tex],-2-[tex]\sqrt{}5[/tex]

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.
 
  • #10
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.
 
  • #11
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.
 

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

What does "ln" stand for in the equation?

"ln" stands for natural logarithm, which is a mathematical function used to calculate the logarithm of a number with base e (approximately 2.718). It is commonly denoted as ln(x) or loge(x).

What is the purpose of solving this equation?

The purpose of solving this equation is to find the value of x that satisfies the equation. This can help in solving various problems in mathematics, science, and engineering that involve exponential or logarithmic functions.

How do I solve this equation?

To solve this equation, you can use the properties of logarithms to simplify it. First, combine the logarithms on the left side by using the quotient rule, ln(x+2)-ln(x+1) = ln[(x+2)/(x+1)]. Then, use the fact that ln(a) = 1 is equivalent to a = e, where e is the base of the natural logarithm. This will give you a quadratic equation to solve for x.

Are there any restrictions on the values of x in this equation?

Yes, there are restrictions on the values of x in this equation. Since the natural logarithm of a negative number is undefined, the values of x+2 and x+1 must be positive. This means that x must be greater than -1.

Can this equation have more than one solution?

Yes, this equation can have more than one solution. In fact, it can have two solutions, as it is a quadratic equation. However, it is also possible for the equation to have no real solutions if the discriminant is negative.

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