physics magnification Q?

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assuming a simple magnifier

the image must be twice as far away in order to be twice as large

and the image disdance is negative because the imabe is on the same side of lens as the object

so

1/f = 1/3.0 + 1/(-6.0) = 1/6

f = 6.0 cm

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(1/u)+(1/v)=(1/f)

(1/3)+(1/v(2*3))=(1/f)

(6+3)/(6*3)

9/27=1/3=1/f

f=3