Logarithim question decreasing in value

[pinnacleprouk]

Logarithim question "decreasing in value"

Homework Statement

The Radioactivity (R) of a substance can be modeled using the equation:

R = A x 2 ^ -Bt

Where R is measured in becquerels per gram, t is measured in hours and A and B are constants.
If a substance has an initial radioactivity of 60 becquerels and one hour later it is 50 becquerels find A and B and the time for the radioactivity to reduce to half its initial value (Half Life)

Homework Equations

R = A x 2 ^ -Bt

The Attempt at a Solution

R = A x 2 ^ - Bt

R = 60 t = 0
R = 50 t = 1
60 = A x 2 ^ - Bx0

60 = A

50 = 60 x 2 ^ - B

2 - B = 50/60

2 - B = 0.8

log 2 - B = log 0.8

- B log 2 = log 0.8

- B = log 0.8/log 2

- B = - 0.32

Is this all right?

I work out half life to be around 3 hours but could anybody give me a more exact time?

Thanks in advance

 

[pinnacleprouk]

Solved t = 3.125

Thanks

 

[Mark44]

Homework Statement

The Radioactivity (R) of a substance can be modeled using the equation:

R = A x 2 ^ -Bt

Where R is measured in becquerels per gram, t is measured in hours and A and B are constants.
If a substance has an initial radioactivity of 60 becquerels and one hour later it is 50 becquerels find A and B and the time for the radioactivity to reduce to half its initial value (Half Life)

Homework Equations

R = A x 2 ^ -Bt

The Attempt at a Solution

R = A x 2 ^ - Bt

R = 60 t = 0
R = 50 t = 1
60 = A x 2 ^ - Bx0

60 = A

50 = 60 x 2 ^ - B

The next line is incorrect. It should be 2^-B = 50/60 = 5/6

2 - B = 50/60

The next line is also incorrect. 5/6 != .8

2 - B = 0.8

log 2 - B = log 0.8

- B log 2 = log 0.8

The next line is incorrect, but at least I understand what you are doing. It would be correct if you had on the right side (log 5/6)/log 2

- B = log 0.8/log 2

- B = - 0.32

You're actually off by quite a bit. To 4 decimal places, B = .2630

Is this all right?


I work out half life to be around 3 hours but could anybody give me a more exact time?

It's about 3.8 hours.

Thanks in advance