# Half life for Carbon-14 is 5,730 years. If yous tart out with 50 grams, how long will it take until it reaches 2 grams? Use A = P(1/2)^t/5730 Find t correct to the nearest whole year.

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### 3 Answers

`A=P(1/2)^(t/5730)`

To solve for the number of years it takes to the carbon to decay from 50g to 2g, plug-in A=2 and P=50 to the given formula.

`2=50(1/2)^(t/5730)`

Then, divide both sides by 50 to have (1/2)^(t/5730) at the right side of the equation.

`2/50=(50(1/2)^(t/5730))/50`

`1/25=(1/2)^(t/5730)`

To remove the t in the exponent, apply the loagrithm property ln `a^m = mlna` . So, take the natural logarithm of both sides of the equation.

`ln(1/25)=ln(1/2)^(t/5730)`

`ln(1/25)=t/5730ln(1/2)`

Then, isolate t. To do so, multiply both sides by 5730.

`5730*ln(1/25) = t/5730 ln(1/2) * 5730`

`5730ln(1/25)=tln(1/2)`

And, divide both sides by ln(1/2).

`(5730ln(1/25))/(ln(1/2))=(tln(1/2))/(ln(1/2))`

`(5730ln(1/25))/(ln(1/2))=t`

`26609.296=t`

Rounding off to nearest whole number, it becomes:

`t=22609`

**Hence, it takes 22,609 years for the Carbon-14 to decay from 50g to 2g.**

The half life of carbon 14 is 5730 years. If the starting amount of C-14 is P, the amount after t years is given by `A = P*(1/2)^(t/5730)`

Let the time taken for 50 g to decay to 2 g be t:

2 = `50*(1/2)^(t/5730)`

=> `log(1/25) = (t/5730)*log(1/2)`

=> `t = 5730*(log(1/25))/(log(1/2))`

=> t `~~` 26609 years

**The time in which 50 g of C-14 decays to 2 g is approximately 26609 years.**

Hi, Kristin.

First, we sub in 50 for P and 2 for A. Then, we solve for t.

First, divide each side by 50. So:

2/50 = (1/2)^t/5730

Then, take the log (base 10) of each side:

log(2/50) = log((1/2)^t/5730)

With this, though, the exponent on the right comes down in front of the log, as a factor, a multiplier. So, we have:

log(2/50) = (t/5730)*log(1/2)

Then, we can divide each side by log(1/2)"

[log(2/50)]/[log(1/2)] = t/5730

The right side is 4.644. Then, multiply each side by 5730, giving t = 26,610.12. So, it would take 26,610.12 years to get 2 grams.

I hope this helps, Kristen. Good luck.

Till then,

Steve