Answer precalculus word problem.

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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

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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.

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