Radioactive decay - log question

[karmatic]

Homework Statement

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?

Homework Equations

x^{\frac{m}{n}}=(x^{\frac{1}{n}})^{m}

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?

 

[Dick]

Homework Statement

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?

Homework Equations

x^{\frac{m}{n}}=(x^{\frac{1}{n}})^{m}

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?

That looks just fine to me.

 

[karmatic]

That looks just fine to me.

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

 

[Curious3141]

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

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').