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Okay I'm almost there but the answer I'm getting is incorrect.

Mathematics
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\[\ln (x)+\ln (x-2)=\ln 8\]
then...\[\ln (x)(x-2)=\ln 8\]
then...\[\ln (x ^{2}-2x)=\ln 8\]

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Other answers:

then drop the ln... \[x^{2}-2x=8\]
going good...
well you e^ both sides. That's how the ln cancel. But then you are left with a quadratic equation
split the two, \[x^{2}=8\] and \[-2x=8\]
thats incorrect
true ok ok
u need quadratic formula
yup
Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
x^2-2x-8=0
okay gimme a second to work that out.
sure.
*** and need to remember to check for valid values of x. if you are taking ln(x) and ln(x-2) what can't x be?
You can always complete the square. That's another way of solving this without the long work of using the quadratic
2
whats finishing the square? Sounds like a shortcut I was denied...
granted
shortcut only if u have some prsctice.... u wanna learn that ?
... or just factor? \[x^2-2x-8=0=(x-?)(x+?)\]
practice some more then use that form
factoring is best here..
can u factor ?
I'll give it a shot! :)
sure do :)
go han!!!! go ku!!!! dominate the problem TEAMWORKKKK
? jerry ?
GO HAN GO KU
\[x^2−2x−8=0=(x−A)(x+B)\\ A*B=-8\\ A+B=-2\] Find A and B
Almost there....
You have a system of question where you can simply plug in one equation to another if you can't see it right away. Then solve for the missing variable.
its (x-4)(x+2)
thats correct :)
so what are 2 values of x u got ?
ten solving its 4 and -2
\[(x-4)(x+2)=0\]
and -2 would be extraneous
but can x be -2 ?? think ....
yes, so x is only = ?
4?
yup, u got it :)
HAZUAHH!
THanks everyone! I really appreciate it! :)
welcome ^_^

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