help with logarithms?

👨🏻‍🏫 學術台數學

[匿名]

2008-12-13 7:39 am

Using suitable substitutions, solve this equation:
2e^x=3-e^(x+1)

更新1:

thanks a lot, but i'll give the best answer to whoever's answer actually HAS a substitution in it.

回答 (4)

ted s

2008-12-13 7:48 am

✔ 最佳答案

you have 2 e^x = 3 - e*e^x---> [2+e] e^x = 3---> x = ln[3/(2+e)] = ln 3 - ln(2+e)

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[匿名]

2008-12-13 3:51 pm

2e^x = 3 - eÂ¹e^x

2(e^x) + ee^x = 3

(e^x) (2 + e) = 3

e^x = 3 / (2+e)

ln(e^x) = ln(3 / (2+e))

x = ln(3/ (2+e)) â -0.4528

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An ESL Learner

2008-12-13 6:46 pm

2e^x = 3 - e^(x + 1)
2e^x + e^(x + 1) = 3
e^x(2 + e^1) = 3
e^x(2 + e) = 3
e^x = 3/(e + 2)
x = ln[3/(e + 2)]

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[匿名]

2008-12-13 3:52 pm

2e^x = 3 - e^(x+1)
2e^x + e^(x+1) = 3
e^x = 3/(2+e)
x = ln(3/(2+e))

Answer: x = ln(3/(2+e)) ... â -0.45283

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