Suppose the half-life of a certain radioactive substance is 20 days and there are 50g present initially. a)
Write an equation that models this situation.
b)
How much radioactive substance is remaining after 12 days?
c)
Find the time when there will be 1g of the substance remaining.

a)   10(5g + 2r)   weight loss over time

b)   20/12 = 1.66667 = 50 / 1.67 = 29.999 gm = 5(6g -1/50)  or  5(5 + g) = 30g

c) 30g = 12 days = 3/5   therefore 5/5 -3/5 = 2/5  = 2/5 x 50 (g) = 20g in 8 days lost

8/20 = 0.4  loss each day 50g - 0.8g = 20-2 = 18 days.

Step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

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step-by-step explanation: