Answer precalculus word problem.

Mathematics
Tutor: None Selected Time limit: 1 Day

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

answer step by step 

Aug 30th, 2015

Thank you for the opportunity to help you with your question!

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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