SImple Logarithms

[anil]

I need help on a lograrith HM:

Solve for X: log(x-1)=log(x-2)-log(x+2)
I got till here: log(x-1)=(log(x-2)/(x+2))

Donot know what to do after wards

***Hey if you could tell me the answer please tell me how you derived it. i want to understand the problem. And Base if course to the 10.

[Tom Mattson]

Originally posted by anil
I need help on a lograrith HM:

Solve for X: log(x-1)=log(x-2)-log(x+2)

***Hey if you could tell me the answer please tell me how you derived it. i want to understand the problem. And Base if course to the 10.

Hi anil,

I'll give you a hint that should unlock the whole problem for you.

log(a)-log(b)=log(a/b)

You can apply that to the right side, and then take the antilog of both sides.

[Cummings]

Originally posted by Tom
Hi anil,

I'll give you a hint that should unlock the whole problem for you.

log(a)-log(b)=log(a/b)

You can apply that to the right side, and then take the antilog of both sides.
:) ah tom, he has already done that
log(x-1)=(log(x-2)/(x+2))
its after that he is stuck :)

to antilog:
Since the bases are the same (log e), raise both sides to the power of e. This will eliminate the log e on both sides leaving you with a simple equation to solve.

[HallsofIvy]

To Tom: As Cummings pointed out, the anil had already done the
log(x-1)= log((x-2)/(x+2)) part.

To Cummings: anil also said "And Base if course to the 10."!

Actually it doesn't matter: as long as a function is "one-to-one" (and logarithm to any base is one-to-one) if f(x)= f(y) then x= y.

Since log(x-1)= log((x-2)/(x+2)) we have x-1= (x-2)/(x+2). Now multiply both sides by x+2 to get (x+2)(x-1)= x-2 so you have the quadratic equation x²+ x- 2= x-2. You ought to be able to solve that easily!

[anil]

Actually i solved it like this:

log(x-1) = log(x-2)-log(x+2)
log(x-1) = log((x-2)/log(x+2))
10^log(x-1) = 10^log((x-2)/(x+2))
(x-1) = (x-2)/(x+2)
(x-1)(x+2) = (x-2)(x+2)/(x+2)
x^2+2x-x-2 = (x-2)
x^2+(x-2) = (x-2)
x^2+(x-2)-(x-2) = (x-2)-(x-2)
x^2 = 0
sqrt(x^2) = sqrt(0)
x = 0
But if you Plug o back in Log(0-1) = not possible in real values? So thats is an extraneous solution?

[Tom Mattson]

Originally posted by Cummings
:) ah tom, he has already done that
log(x-1)=(log(x-2)/(x+2))
its after that he is stuck :)


Originally posted by HallsofIvy
To Tom: As Cummings pointed out, the anil had already done the
log(x-1)= log((x-2)/(x+2)) part.


No fair, he edited that part in as I made my comment. [:(]
(See how his edit and my post are 1 second apart?)

[anil]

I posted it and then i was reading the rules. So it says that post it so that people know you tried. SO i didn't read the post you posted. SOrry tom =(

[HallsofIvy]

Okay, Tom, I see now.

(You know it has never occurred to me to go back and edit my mistakes! I'll have to start doing that (and then pretend I didn't make a mistake from the start!)

Anil, you are correct that the equation log(x-1)= log(x+2)- log(x+2) has NO real number solutions.