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A late is stocked with 1500 trout, the # of original trout alive after x yrs is given by T(x)=1,500e^-1.4x When will there be 300 of original trout left

Aug 30th, 2015

We must solve the equation 1500e^(-1.4x) = 300. Dividing both sides by 1500, we get e^(-1.4x) = 1/5.

The LHS can be rewritten as 1/e^(1.4x), so we have 1/e^(1.4x) = 1/5.

Multiplying both sides by e^(1.4x) * 5, we get 5 = e^(1.4x).

Now take the natural logarithm of both sides: ln(5) = ln(e^(1.4x)). Because, in general, ln(e^z) = z, we get

ln(5) = 1.4x, so x = ln(5)/1.4 ~ 1.609/1.4 = 1.15 years.

CHECK: 1500 * e^(-1.4 * 1.15) = 1500 * e^(-1.61) = 1500/5.0028 ~ 300.

You might find the following useful: http://mathonweb.com/help_ebook/html/functions_3.htm Please let me know if you need any clarification. I'm always happy to answer your questions.
Aug 30th, 2015

sorry, i meant that  the equation was T(x)=1,500e^-0.4x

Aug 30th, 2015

OK. In my opinion, you would be better off trying to do the problem based on the steps I showed, replacing each occurrence of 1.4 with 0.4 and recalculating, than having me do it. If you have trouble doing so, let me know.

Aug 30th, 2015

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Aug 30th, 2015
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Aug 30th, 2015
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