[Help] Very hard maths question;

The post was closed by under_kid-as

under_kid-as — Wed, 01 Jul 2009 21:41:34 +0100

Reply from under_kid-as

under_kid-as — Wed, 01 Jul 2009 21:40:36 +0100

Thank you (: Very much appreciated. x

Reply from Usernam

Usernam — Wed, 01 Jul 2009 21:39:24 +0100

Here’s something else i found btw none of these are my answeres i found them on different websites, this problem is almost the same,:

ln(x) + ln(x - 3) = 0

To solve this, first combine the logs using the property that the sum of logs is equal to the log of a product.

ln[ x(x - 3) ] = 0

Convert to exponential form.

e^0 = x(x - 3)

And solve for x.

1 = x(x - 3)
1 = x^2 - 3x
0 = x^2 - 3x - 1

Therefore,

x = [ 3 +/- sqrt(9 - 4(-1)) ] / 2
x = [ 3 +/- sqrt(13) ] / 2

However, with logarithmic equations, not all evaluated solutions may necessarily work. We must plug each solution into the original.
Note that sqrt(13) lies somewhere between 3 and 4, because
9 13 16, so
sqrt(9) sqrt(13) sqrt(16), or
3 sqrt(13) 4

This means 3 - sqrt(13) is a negative number, meaning
[3 - sqrt(13)]/2 is also a negative number. We cannot take the log of a negative number, and, as per the original question,

we have ln(x -3), and subtracting 3 from an already negative number is negative. Therefore, we reject x = [3 - sqrt(13)]/2 as a possible solution, which means

x = [ 3 + sqrt(13)]/2

Reply from Usernam

Usernam — Wed, 01 Jul 2009 21:32:50 +0100

e^(lnx)=-e^(-ln(x-3))

x=1/(x-3)

x=3.30278

Reply from under_kid-as

under_kid-as — Wed, 01 Jul 2009 21:30:51 +0100

[quote Usernam]ln[x(x-3)] = ln1
x(x-3) = 1
Collecting all terms in one side,
x^2 - 3x - 1 = 0
Using quadratic formula with x > 3, (the domain of ln(x-3))
x = [3+sqrt(13)]/2[/quote]

can you do that with the logs?!?!?

Reply from Usernam

Usernam — Wed, 01 Jul 2009 21:29:19 +0100

ln[x(x-3)] = ln1
x(x-3) = 1
Collecting all terms in one side,
x^2 - 3x - 1 = 0
Using quadratic formula with x > 3, (the domain of ln(x-3))
x = [3+sqrt(13)]/2