Write the following expression as a sum and/or difference of the logarithms.
log((sqrt(x+2) / x^2-x))
I know the rules....I just need the answer ASAP...I thought I knew the answer, but it was wrong and I don't know why? Please help me....I promise to give 10 to best answer!
2007-12-05 02:52:31 · 6 answers · asked by ☺♠JonasJay♫♦ 5 in Science & Mathematics ➔ Mathematics
I think the answer is (1/2 log (x+2)) - log x - log (x-2)........you all agree that just answered or anyone? I think my expression is written correctly...maybe it is.... log (all this next part is in one big parethesis: sqrt(x+2)/x^2-x
2007-12-05 03:05:47 · update #1
log [√(x+2) / (x² - x)] =
log √(x+2) - log (x² - x) =
(1/2) log (x+2) - log (x² - x)
if that's how you intended the ().
2007-12-05 02:59:31 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
Is log((sqrt(x+2) / x^2-x))
by any chance really supposed to be
log(sqrt(x+2) /( x^2-x))?
2007-12-05 02:56:45 · answer #2 · answered by Steve H 5 · 0⤊ 0⤋
log(four x square minuous log x)
Sorry I can't find no symbols. I tryed to my best of my knowledge
2007-12-05 03:02:52 · answer #3 · answered by robles_flores_298@sbcglobal.net 1 · 0⤊ 0⤋
log(a/b) = log a - log b, so:
log (sqrt(x+2)) - log (x^2-x)
2007-12-05 02:56:38 · answer #4 · answered by grompfet 5 · 0⤊ 0⤋
if the circle with radius r, has center (a,b) on the coordinate airplane the equation might want to be (x-a)^2+(y-b)^2=r^2 so for the equation (x^2)+6x+(y^2)-2y=2 attempt to "rework" it into usual kind (x^2)+6x+(y^2)-2y=2 upload 10 to each and each side to finish the sq. (x^2)+6x+9+(y^2)-2y+a million=2+9+a million <=> (x+3)^2 +(y-a million)^2=12 in usual kind center is (3,-a million) and r=?12=2?3 for the hyperbola (x^2)-4x-(y^2)+6y=9 finished the squares back (x^2)-4x-(y^2)+6y=9 (x^2)-4x+4-(y^2)+6y-9=9-9+4 (x-2)^2-(y-3)^2=4 Dividing each and each side by ability of four ((x-2)^2)/4-((y-3)^2)/4)=a million ((x-2)^2)/2^2-((y-3)^2)/2^2)=a million this is contained in the usual kind
2016-10-25 12:02:45 · answer #5 · answered by ? 4 · 0⤊ 0⤋
235743895749573953756xycubed
2007-12-05 02:55:53 · answer #6 · answered by Anonymous · 0⤊ 0⤋