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Mathematics 20 Online
OpenStudy (anonymous):

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

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...

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

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.

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

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 ?

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

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 :)

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 ....

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! :)

OpenStudy (anonymous):

welcome ^_^

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