3. the equation for radioactive decay is $p=(0.5)^{\frac{t}{h}}$, where $p$ is the part of a substance with…

3. the equation for radioactive decay is $p=(0.5)^{\frac{t}{h}}$, where $p$ is the part of a substance with half - life $h$ remaining radioactive after a period of time, $t$. a given substance has a half - life of 6,000 years. after $t$ years, one - fifth of the original sample remains radioactive. find $t$, to the nearest thousand years.

3. the equation for radioactive decay is $p=(0.5)^{\frac{t}{h}}$, where $p$ is the part of a substance with half - life $h$ remaining radioactive after a period of time, $t$. a given substance has a half - life of 6,000 years. after $t$ years, one - fifth of the original sample remains radioactive. find $t$, to the nearest thousand years.

Answer

Explanation:

Step1: Substitute given values into formula

We know that $p=\frac{1}{5}$, $H = 6000$. Substitute into $p=(0.5)^{\frac{t}{H}}$, getting $\frac{1}{5}=(0.5)^{\frac{t}{6000}}$.

Step2: Take the natural - logarithm of both sides

$\ln(\frac{1}{5})=\ln((0.5)^{\frac{t}{6000}})$. Using the property $\ln(a^b)=b\ln(a)$, we have $\ln(\frac{1}{5})=\frac{t}{6000}\ln(0.5)$.

Step3: Solve for $t$

First, $\ln(\frac{1}{5})=-\ln(5)\approx - 1.6094$ and $\ln(0.5)\approx-0.6931$. Then $t = 6000\times\frac{\ln(\frac{1}{5})}{\ln(0.5)}$. Substitute the values of the logarithms: $t = 6000\times\frac{-1.6094}{-0.6931}\approx13900$. Rounding to the nearest thousand, $t\approx14000$.

Answer:

$14000$