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- #1

- Thread starter goosey00
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- Thread starter
- #1

- Mar 1, 2012

- 249

You can check by evaluating $e^{-0.38*0.4387}$Solve for x:e^-0.38x=.3

I got .4387 Is that correct

If we use google calculator we end up with 0.8467 (4sf) so 0.4387 is not correct.

What do you know about solving exponential equations and/or the natural logarithm?

- Thread starter
- #3

- Mar 1, 2012

- 249

Either

Code:

` [2nd] [ln] [(] [-][0.38] [x] [0.4387][)][=]`

Code:

` [(] [-][0.38] [x] [0.4387][)][2nd] [ln][=]`

Bear in mind that was just a test to see if your answer was right (it isn't). You need to use the natural logarithm (ln) to find x.

$-0.38\ln(x) = ln(0.3)$

Solve for [tex]x:\;e^{-0.38x}\:=\:0.3[/tex]

I got 0.4387 . Is that correct?

Can't you check your answer?

[tex]\text{We have: }\:e^{-0.38x} \;=\;0.3[/tex]

[tex]\text{Take logs: }\:\ln(e^{-0.38x}) \;=\;\ln(0.3) \quad\Rightarrow\quad \text{-}0.38x\underbrace{\ln e}_{\text{This is 1}} \;=\;\ln(0.3)[/tex]

. . . [tex]\text{-}0.38x \;=\;\ln(0.3) \quad\Rightarrow\quad x \;=\;\frac{\ln(0.3)}{\text{-}0.38}[/tex]

. . . . . [tex]x \;=\;3.168\,349\,485[/tex]