Answer to Question #2630 in Algebra for Azrael Linx
I've tried to answer this question about ten times, not joke. I don't seem to understand where I'm going wrong. If you could just explain the method behind it, I'd be glad.
3e^x + 6e^-1 - 11 = 0
It seems to be exponential equation if the equation is written correctly: 3ex = 11-6/e ex = (11-6/e)/3 x = ln ((11 - 6/e)/3) ≈ 1.075