# 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

Thanks.

3e^x + 6e^-1 - 11 = 0

Thanks.

Expert's answer

It seems to be exponential equation if the equation is written correctly:

3e

e

x = ln ((11 - 6/e)/3) ≈ 1.075

3e

^{x}= 11-6/ee

^{x}= (11-6/e)/3x = ln ((11 - 6/e)/3) ≈ 1.075

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