# Radioactive decay - log question

1. Apr 27, 2013

### karmatic

1. The problem statement, all variables and given/known data

The amount (A) of cesium-137 remaining after t years is given by

A=A$_{0}$*2$^{\frac{-t}{30.3}}$

where A$_{0}$ is the initial amount. In what year will the cesium-137 be 10% of that which was released at the Chernobyl disaster in 1986?

2. Relevant equations
x$^{\frac{m}{n}}$=(x$^{\frac{1}{n}}$)$^{m}$

3. The attempt at a solution
plug in my 0.1 on the LHS

0.1A$_{0}$=A$_{0}$*2$^{\frac{-t}{30.3}}$

A$_{0}$ on both sides cancels out

0.1=2$^{\frac{-t}{30.3}}$

convert decimal to a fraction

$\frac{1}{10}$=2$^{\frac{-t}{30.3}}$

take the reciprocal of the RHS to convert the negative power to a positive

$\frac{1}{10}$=$\frac{1}{2^{\frac{t}{30.3}}}$

take the reciprocal of both sides to remove the fractions

10=2$^{\frac{t}{30.3}}$

isolate the variable t and calculate the remaining values

10=(2$^{\frac{1}{30.3}}$)$^{t}$

10=1.02313981$^{t}$

then this part I'm unsure if I'm doing right, and I'm not aware of how I should be trying to check my work

t=$\frac{log 10}{log 1.02313981}$

t=100.6544213

1986+100=2086?

Last edited: Apr 28, 2013
2. Apr 28, 2013

### Dick

That looks just fine to me.

3. Apr 28, 2013

### karmatic

thank you very much for the quick response, my mind has been put at ease

4. Apr 28, 2013

### Curious3141

For future reference, there's a neater and less error-prone way to do the manipulation. Basically, we're using a rule of logs: $\log a^b = b\log a$, where the logarithm can be to any base.

In this case, use common (base 10 logs) because one side is a power of 10 (actually, 10 itself). So:

$$10 = 2^{\frac{t}{30.3}}$$

Take common logs of both side:

$$1 = \log_{10}2^{\frac{t}{30.3}}$$

Use that rule mentioned above:

$$1 = {\frac{t}{30.3}}\log_{10}2$$

Rearrage:

$$t = \frac{30.3}{\log_{10}2} = 100.65$$

and so forth. The calculation can be done in one step right at the end. Most modern scientific calculators can immediately calculate base-10 logs (it's the button marked 'log').