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 Post subject: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 19:30:32 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
Posts: 18
Hi,

Could someone tell me the correct way to use ln to solve this equation. I keep getting it wrong.

I am trying to find the answer to solve for i

650 = 300 (1 + i)^8.69231

Last edited by nquadr on Mon, 6 Feb 2012 19:56:05 UTC, edited 1 time in total.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 19:33:58 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
Hi,

Could someone tell me the correct way to use ln to solve this equation. I keep getting it wrong.

I am trying to find the answer to solve for i

650 = 350 (1 + i)^8.69231

Divide both sides by 350 and take the log of both sides.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 19:46:52 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
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This is what I have done so far, but it is not the right answer, answer should be .093027

ln(650/300) / ln (8.69231) this should leave me with 1+ i

and to solve for i, I subtract by 1 and get -.713731 instead of the correct answer above.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 19:50:43 UTC
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This is what I have done so far, but it is not the right answer, answer should be .093027

ln(650/300) / ln (8.69231) this should leave me with 1+ i

and to solve for i, I subtract by 1 and get -.713731 instead of the correct answer above.

First of all, why do you have ? Shouldn't that be ?

If that's the case, then notice when you log both sides you get:

.

Try taking it from that point.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 20:02:04 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
Posts: 18
Sorry it should be 300, so far I have

ln(650/300)/8.69231 = ln(1+i)

or

.088951 = ln (1+i)

I am not sure how to evaluate ln (1+i), I know ln (1) = 0; but that would not make sense.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 20:28:19 UTC
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Joined: Mon, 6 Feb 2012 05:44:39 UTC
Posts: 27
now exponentiate both sides to get rid of ln

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 20:28:32 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
Posts: 18
Ok I think I got it, raise the value as an exponent with base e and subtract 1, gets me the correct answer. So I will assume it is correct. Thanks.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 20:29:45 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
Posts: 18
Thank you.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 22:15:16 UTC
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Ok I think I got it, raise the value as an exponent with base e and subtract 1, gets me the correct answer. So I will assume it is correct. Thanks.

Yes.

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 Post subject: Re: Using the ln to find the a variablePosted: Mon, 6 Feb 2012 22:23:47 UTC
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650 = 300 (1 + i)^8.69231

(1 + i)^8.69231 = 650/300 = 13/6

i = (13/6)^(1/8.69231) - 1 = ~.09302712

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