Question

to find the time at which only 1 mg remains, we must solve 1 = y(t) = 40(2−t/30), and so we get the following. t = −30 log2

Answer 1

To find the time at which only 1 mg remains, we need to solve the equation
`\displaystyle\sf 1 = y(t) = 40(2-(t)/(30))`, where
`\displaystyle\sf t` represents time.

Let's solve for
`\displaystyle\sf t`:

`\displaystyle\sf 1 = 40(2-(t)/(30))`.

Dividing both sides of the equation by 40:

`\displaystyle\sf (1)/(40) = 2-(t)/(30)`.

Subtracting 2 from both sides:

`\displaystyle\sf -(79)/(40) = -(t)/(30)`.

Multiplying both sides by 30:

`\displaystyle\sf -(79)/(40) * 30 = -t`.

Simplifying:

`\displaystyle\sf t = -30 \log 2`.

Therefore, the time at which only 1 mg remains is
`\displaystyle\sf t = -30 \log 2`.

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