If $\displaystyle \frac{16^x}{\frac{1}{2} x^2}-32=0$, then $\displaystyle x=$

(a) -1 or -4

(b) 1 or -5

(c) 2 or -4

(d) -2 or 4

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- Dec 9th 2009, 02:13 PMibnashrafmultiple choice question
If $\displaystyle \frac{16^x}{\frac{1}{2} x^2}-32=0$, then $\displaystyle x=$

(a) -1 or -4

(b) 1 or -5

(c) 2 or -4

(d) -2 or 4 - Dec 9th 2009, 02:59 PMbigwave
- Dec 9th 2009, 03:07 PMProve It
I would advise graphing this equation and seeing where they cross the $\displaystyle x$ axis.

Equivalently, you could graph

$\displaystyle y = 16^x$ and $\displaystyle 16x^2$ and see where they intersect. - Dec 9th 2009, 03:27 PMibnashraf
actually that's how i interpreted the question to be since it wasn't clearly written. This is exactly how i saw it written down:

$\displaystyle 16^x/\frac{1}{2}x^2-32=0$

So is there any way that the above question can be "interpreted" to give one of the 4 responses mentioned previously? - Dec 9th 2009, 03:53 PMbigwave
as Prove IT suggested to graph it

did graph this and none of answers showed except x = 1 is the only interger value

and got 1/2

its a strange multiple choice - Dec 9th 2009, 03:54 PMANDS!