A laboratory has 20 grams of a radioactive substance that decays in such a way that t days later the number of grams left will be y = 20 (1/10)t/15. How many days does it take for half of the substance to decay?
A. 4
B. 4.5
C. 9
D. 9.5

Question

A laboratory has 20 grams of a radioactive substance that decays in such a way that t days later the number of grams left will be y = 20 (1/10)t/15. How many days does it take for half of the substance to decay? 
	A.	4
	B.	4.5
	C.	9
	D.	9.5

whpalmer4 · Answer

$$y = 20(\frac{1}{10})^{t/15}$$
To find the value of $t$ such that $y(t) = 10$, just plug in the numbers and solve:

$$10 = 20(0.1)^{t/15}$$Divide both sides by 20$$0.5=(0.1)^{t/15}$$Remember the property of logarithms that $\log a^b = b \log a$ and take the log of both sides:
$$\log 0.5 = \frac{t}{15} \log 0.1$$I think you should be able to solve it from there.

whpalmer4 · Answer

I find it especially worthwhile to plug the answer back into the original formula to check the answer when doing problems like this!