Okay I'm almost there but the answer I'm getting is incorrect.
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OpenStudy (anonymous):
\[\ln (x)+\ln (x-2)=\ln 8\]
OpenStudy (anonymous):
then...\[\ln (x)(x-2)=\ln 8\]
OpenStudy (anonymous):
then...\[\ln (x ^{2}-2x)=\ln 8\]
OpenStudy (anonymous):
then drop the ln...
\[x^{2}-2x=8\]
OpenStudy (anonymous):
going good...
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OpenStudy (anonymous):
well you e^ both sides. That's how the ln cancel. But then you are left with a quadratic equation
OpenStudy (anonymous):
split the two,
\[x^{2}=8\]
and
\[-2x=8\]
OpenStudy (anonymous):
thats incorrect
OpenStudy (anonymous):
true ok ok
OpenStudy (anonymous):
u need quadratic formula
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OpenStudy (anonymous):
yup
OpenStudy (anonymous):
Compare your quadratic equation with \(ax^2+bx+c=0\)
find a,b,c
then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
OpenStudy (anonymous):
x^2-2x-8=0
OpenStudy (anonymous):
okay gimme a second to work that out.
OpenStudy (anonymous):
sure.
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OpenStudy (anonymous):
*** 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?
OpenStudy (anonymous):
You can always complete the square. That's another way of solving this without the long work of using the quadratic
OpenStudy (anonymous):
2
OpenStudy (anonymous):
whats finishing the square? Sounds like a shortcut I was denied...
OpenStudy (anonymous):
granted
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OpenStudy (anonymous):
shortcut only if u have some prsctice.... u wanna learn that ?
OpenStudy (anonymous):
... or just factor? \[x^2-2x-8=0=(x-?)(x+?)\]
OpenStudy (anonymous):
practice some more then use that form
OpenStudy (anonymous):
factoring is best here..
OpenStudy (anonymous):
can u factor ?
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OpenStudy (anonymous):
I'll give it a shot! :)
OpenStudy (anonymous):
sure do :)
OpenStudy (anonymous):
go han!!!! go ku!!!! dominate the problem TEAMWORKKKK
OpenStudy (anonymous):
? jerry ?
OpenStudy (anonymous):
GO HAN GO KU
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OpenStudy (anonymous):
\[x^2−2x−8=0=(x−A)(x+B)\\ A*B=-8\\ A+B=-2\]
Find A and B
OpenStudy (anonymous):
Almost there....
OpenStudy (anonymous):
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.
OpenStudy (anonymous):
its (x-4)(x+2)
OpenStudy (anonymous):
thats correct :)
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OpenStudy (anonymous):
so what are 2 values of x u got ?
OpenStudy (anonymous):
ten solving its 4 and -2
OpenStudy (anonymous):
\[(x-4)(x+2)=0\]
OpenStudy (anonymous):
and -2 would be extraneous
OpenStudy (anonymous):
but can x be -2 ?? think ....
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OpenStudy (anonymous):
yes, so x is only = ?
OpenStudy (anonymous):
4?
OpenStudy (anonymous):
yup, u got it :)
OpenStudy (anonymous):
HAZUAHH!
OpenStudy (anonymous):
THanks everyone! I really appreciate it! :)
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