MHB Need help solving exponential problem

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The discussion centers on solving an exponential problem related to loan repayment. The user seeks assistance with a loan of $10,000 at a 2.14% interest rate, compounded monthly, with monthly payments of $230. The solution involves approximating the time to repayment using the Newton-Raphson method, which was suggested by another forum member. The user is encouraged to provide the complete problem for more tailored advice. Understanding the full context is crucial for effective assistance in solving the exponential equation.
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Hi,

I'm new to this forum so thank you in advance for any help on this problem. I would like to understand the steps needed to solve this problem. The answer is 3.94111..286 = (1.00178^(12 * t))/t or .286 = (e^.021341 * t)/t
 
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Re: need help solving exponential problem

Hi arrow, welcome to MHB!:)

In your question, I believe there's really no easy way(that I know of) to do it, but one can approximate its only solution by using the Newton-Raphson method, have you heard of this method before?
 
Re: need help solving exponential problem

Hi Anemone,

No, I'm not familiar with this method. I'll take a look. Thanks.
 
Re: need help solving exponential problem

arrow said:
Hi Anemone,

No, I'm not familiar with this method. I'll take a look. Thanks.

Oh okay...but would it be possible for you to restate the problem in its entirety here? We might be able to provide other useful advice to you if you post the full question for us first...
 
Re: need help solving exponential problem

The question was this. A person is taking out a loan of $10,000, compounded monthly, with an interest rate of 2.14 percent. She is making payments of $230 / mo. How long before the loan is repaid?
 
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I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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