Log word question

I have plugged in the numbers but not sure what to do next. y=100e^5700K

Since you have used 100 units of Carbon-14, and you have 1 half-life, it therefore follows that "y" to the left-hand side will be 50 units. But instead of using 5700 as a power of e, I think you should use 1, because 5700=1 half-life. using 5700 as a power of e, you are going to get an extremely large number or an extremely small number, being the reciprocal of 5700. Do you see it? Then just solve for the constant K as follows:

Solve for K:
50 = 100 e^1 K

50 = 100 e K is equivalent to 100 e K = 50:
100 e K = 50

Divide both sides by 100 e:
Answer: | K = 1/(2e)

P.S. There is a much simpler way of finding out the amount of a radioactive material left over after n half-lives: Since you have 1 half-life in this example, then it goes like this:

2^-n X 100=2^-1 X 100=1/2 X 100=50% of the original material left after 1 half-life.

I think this is how you want to to solve for K:

y=yo.e^(K5700), now you set y to the left 1/2y and you get

1/2y=yo.e^(K5700) divide both sides by y

1/2=e^(K5700) take the natural log of both sides

-0.693147=K5700 and

K=-.693147/5700

K=-1.215 X 10^-4 and that's what you are trying to get.

In general, the half life, L,  and the constant k are related by k = -ln(2)/L, so k = -ln(2)/5700 per year = 1.216*10^-4, as Guest has determined in the previous reply.