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Answer to Question #2630 in Algebra for Azrael Linx

Question #2630
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

Expert's answer
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

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