An object is located 51 millimeters from a diverging lens. The object
has a height of 13 millimeters and the image height is 3.5
millimeters. How far in front of the lens is the image located?

The formula is explained on page 46 & 47 of this document. Please let
me know if this answers your question.

http://farside.ph.utexas.edu/teaching/302l/geometric.pdf

tutuzdad-ga

Ok, lets see here:

Essentially, all you need is these equations:

Magnitude = (height of image)/(height of object)

and

Magnitude = - (distance of image)/(distance of object)

Just set these equal to eachother, and you get:

(height of image)/(height of object) = - (distance of image)/(distance of object)

Now, just plug in the numbers you gave, and you get:

3.5/13 = - (distance of image)/51

So, the distance of the image = -51 x (3.5/13) =
-13.73075923076923076923....(repeat the 076923)

(the negative means that its just back towards the original object,
aka, in front of the lens)

You can find these formulas here:
http://www.sparknotes.com/testprep/books/sat2/physics/chapter18section5.rhtml

I just hope this wasn't someone trying to shirk their physics homework....:P