How do you find the associated exponential decay or growth model: Q = 1,500 when t = 0; half-life = 1?

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How do you find the associated exponential decay or growth model: Q = 1,500 when t = 0; half-life = 1?

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Answer 1 · 2017-05-20T19:44:36

Start with $Q (t) = Q (0) e^{λ t}$ $Q(t) = Q(0)e^(lambdat)$
Use the fact that $\frac{Q (t_{half-life})}{Q (0)} = \frac{1}{2}$ $(Q(t_"half-life"))/(Q(0)) = 1/2$ to find $λ$ $lambda$

Explanation:

Start with $Q (t) = Q (0) e^{λ t} [1]$ $Q(t) = Q(0)e^(lambdat)" [1]"$

At $t_{half-life} = 1$ $t_"half-life" = 1$

$\frac{Q (1)}{Q (0)} = \frac{1}{2} = e^{λ (1)}$ $(Q(1))/(Q(0)) = 1/2 = e^(lambda(1))$

Use the natural logarithm on both sides:

$ln (\frac{1}{2}) = ln (e^{λ (1)})$ $ln(1/2) = ln(e^(lambda(1)))$

Flip the equation:

$λ = ln (\frac{1}{2})$ $lambda = ln(1/2)$

This is a better form:

$λ = - ln (2)$ $lambda = -ln(2)$

Substitute $1500$ $1500$ for $Q (0)$ $Q(0)$ and $- ln (2)$ $-ln(2)$ for $λ$ $lambda$ into equation [1]:

$Q (t) = 1500 e^{- ln (2) t}$ $Q(t) = 1500e^(-ln(2)t)$

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How do you find the associated exponential decay or growth model: Q = 1,500 when t = 0; half-life = 1?

1 Answer

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