# exp(y*log(x)) Much Faster than Math.pow(x,y)

Apr 8, 2011 · 1 minute read · CommentsToday I found out that replacing **Math.pow(x,y) **by **Math.exp(y*Math.log(x)) **made me gain 50% performance in my program. Of course, both x and y are double in my case. I find this quite surprising, I expected better from Math.pow.