How do you solve -1 + 8e^(2x-5) = e

[From: ] [author: ] [Date: 12-06-27] [Hit: ]

464785228557 = 2.718281828459 - Solving e^ - 0.766179854162- 1+ 3.718281828459 = 2.718281828459-Multiplication of 0.464785228557 and 8 2.......

-1 + 8*2.718281828459 ^( - 0.766179854162 ) = 2.718281828459 - Subtraction of 4.233820145838 and 5

-1 + 8 * 0.464785228557 = 2.718281828459 - Solving e^ - 0.766179854162

- 1+ 3.718281828459 = 2.718281828459 - Multiplication of 0.464785228557 and 8

2.718281828459 = 2.718281828459 - Addition

It checks and equals

I hope this helps .

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Hello,

-1 + 8.e^(2x - 5) = e
8.e^(2x - 5) = e + 1 →→→ Add 1 on both sides
e^(2x - 5) = (e + 1) / 8 →→→ Divide by 8 on both sides
2x - 5 = ln[(e + 1) / 8] →→→ Take natural logarithm of both sides
2x = 5 + ln[(e + 1)/8] →→→ Add 5 on both sides
x = {5 + ln[(e + 1)/8]} / 2 →→→ Divide both side by 2

Regards,
Dragon.Jade :-)

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-1 + 8e^(2x-5) = e
8e^(2x-5) = e + 1
e^(2x-5) = e/8 + 1/8
2x - 5 = ln(e/8) + ln(1/8)
2x - 5 = ln((e/8) * (1/8))
2x - 5 = ln(e/64)
2x = ln(e/64) + 5
x = [ln(e/64) + 5]/2
x = 0.921

EDIT.

My goodness, what kind of teacher would give you this problem?

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e^(2x-5)=(e+1)---> 2x-5 = ln(e+1) ---> x = ...

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