Help supplying power to LED

Hi all - I have a LED module out of a UKE SL4 that I am trying to use in a different light body. The mechanical fit is fine and the circuit works as desired, but I need some assistance with the powering of the LED. The SL4 is a 4 c-cell light and the new body is a 2 c-cell. The LED is much dimmer in the new body with the 3v disadvantage. The local BatteriesPlus has 3.6v Lithiums that are c-cell sized. Using those would put 7.2 volts to the LED. The question? Is there danger to reliability or life in using the 3.6v batteries in this light? The application is cave diving, so the light MUST work. Other options I have not considered?

I tried to understand the 'LEDs are current devices, not voltage devices' threads, but it's not exactly 101 level stuff. :shrug:

Dont know what the LED in the SL4 is, so here goes:
LEDs work best by being powered by DRIVERS. ie circuits that supply the correct current.
Your SL4 might alrready have this driver, which would be suited to the original voltage ie 4 cells. So 2 cells might not work very well.
I will wait for more knowledgable members to chime in with more specific advie, but I would say for an application like cave diving, dont mess with it unless you know EXACTLY what you are doing!

Is there an easy (or not too hard) way of figuring out which LED module is in the light? Then I should be able to research the actual module rather than the packaged light itself.

From UKE's web site on the SL4, it says there's a circuit of some sort. It's most likely a buck circuit (which takes a high battery voltage and reduces it for the LED), but I don't know how much it can take. A pair of Li-Ions can be up to 8.4V total, and that might be too much for a light designed for 6V max. I mean, it might work, but I wouldn't risk it. Entirely apart from that is the danger associated with Li-Ions, especially with such large cells.

As you mentioned and Mash recommended, such a light MUST work and you probably shouldn't mess with it unless you know exactly what you're doing.