Solved Math issues

1000*1.5^n where N is the amount of times

 

1000 * Math.pow(1.5, x), given x as your variable. Use a long for this.

 

Just make a for loop and do this calculation in there. Something like this:

Code (Text):

double base = 1000;
for(int i = 0; i < x; i++) base *= 1.5;
 

If this is wrong tell me, it's a quick draft.

 

You're not wrong but see the previous replies to see why you're not right either.

 

This isn't the way you would go about it. What @DizMizzer & @Photon suggests would be the appropriate way.

 

Hmhhh it does give the same result though. Is their method faster? http://prntscr.com/ex7akc

 

I would assume so. And easier and more convenient, but I'm checking it now

 

I'm quite sure that their way under the hood also uses a for loop. I could be wrong though.
In my little time as a developer one of the few things I understood is that performance is everything!
Thanks

 

Our way is faster because we aren't storing the variable over and over. This is also basic math not really programming

 

Could be, the thing is this is a more direct approach with indeed more code (which is a bad thing).

 

Test finished. Actually not as expected.

Code (Text):

1 - Result: 1.9175105923288408E8. Operation took 768 nano seconds.
2 - Result: 1.9175105923288408E8. Operation took 13056 nano seconds.

1 = Your method @xTimPugz.
2 = @Photon 's method

Code if you wanna try it yourself.

EDIT: So basically, double as fast. Considerably faster for a computer, but not noticeable to a human eye.

EDIT 2: @xTimPugz take a look at the edit. That's pretty odd. Your method proved to be faster. I just confused the results with my ignorance.

EDIT 3: @DizMizzer , @xTimPugz 's method is actually faster

 

Well yeah I guess that's because under the hood 5 to the power of 5 is just the same as 5 * 5 * 5 *5 *5.

 

Haha, glad you proved them wrong (me, as well)

 

I don't really care about that haha I just want to enjoy my time on the forums and help around where I can hehe. Just thought it'd be helpful.