help would be much appreciated :)

Question

The half-life for thorium-227 is 18.72 days. The amount A (in grams) of thorium-239 after t years for a 10-gram sample is given by
A(t)=10*0.5^(t)/(18.72)

How long will it take before 9 grams of thorium-227 is left in the sample? (Round your answer to the hundredths place.)

days:__________

Answer 1

The first step would be to insert 9 into the equation:

9=10*0.5^(t/18.72).

 

We are trying to solve for t. As such, we need to divide by 10 next:

0.9=0.5^(t/18.72).

 

This leaves us with an exponential function. In order to solve for a variable in the exponent, you will need to use logs. In this case, we will take the natural log (a log with a base e) of both sides:

ln(0.9) = ln(0.5^(t/18.72))

 

One of our rules of logs states that if you are taking the log of a number that is raised to an exponent, that is the same as multiplying that exponent by just the log of the base. In other words log(x^2) is the same thing as 2log(x). So we can bring the exponent down in front of the log:

ln(0.9) = (t/18.72)ln(0.5)

 

Now we divide by the ln(0.5):

ln(0.9)/ln(0.5) = t/18.72

 

And multiply by 18.72:

t=18.72(ln(0.9)/ln(0.5))

 

Type that into a calculator, and you should get 2.85 days.