# Question #6df28

Apr 5, 2016

Let $\tau$ be the half life.

By definition, after the first half life, the amount of radioactive substance should be half of the initial quantity.

Mathematically,

$\frac{A \left(\tau\right)}{A \left(0\right)} = \frac{1}{2}$
or

$\frac{1}{2} A \left(0\right) = A \left(\tau\right)$.

Thus, proceed to solve the above equation for $\tau$.

$\frac{1}{2} \times 450 {e}^{- 0.04 \times 0} = 450 {e}^{- 0.04 \tau}$

$\frac{1}{2} \times {e}^{0} = {e}^{- 0.04 \tau}$

$\frac{1}{2} = {e}^{- 0.04 \tau}$

Take $\ln$ on both sides.

$\ln \left(\frac{1}{2}\right) = \ln \left({e}^{- 0.04 \tau}\right)$

$- \ln \left(2\right) = - 0.04 \tau$

$\tau = \ln \frac{2}{0.04}$

$\approx 17.3$

Apr 5, 2016

Since you are trying to find the half-life of the substance and you know that $A$ represents the final amount of the substance, set $A = 450 \times \frac{1}{2} = 225$.

Setting $A$ to equal $225$ will allow you to find the half-life of the substance since you are assuming $225 g$ (half the amount of the substance) remains.

Thus:

$A \left(t\right) = 450 {e}^{- 0.04 t}$

$225 = 450 {e}^{- 0.04 t}$

From this point on, we solve for $t$, the half-life of the substance.

Start by dividing both sides of the equation by $450$.

$\textcolor{red}{\frac{\textcolor{b l a c k}{225}}{450}} = \textcolor{red}{\frac{\textcolor{b l a c k}{450 {e}^{- 0.04 t}}}{450}}$

$\frac{1}{2} = {e}^{- 0.04 t}$

Since the bases on both sides of the equation are not the same, take the natural logarithm of both sides.

$\ln \left(\frac{1}{2}\right) = \ln \left({e}^{- 0.04 t}\right)$

Using the natural logarithmic property, ${\ln}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\ln}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, the equation simplifies into:

$\ln \left(\frac{1}{2}\right) = - 0.04 t \cdot \ln \left(e\right)$

Solving for $t$, the half-life of the substance is:

$- 0.04 t = \ln \frac{\frac{1}{2}}{\ln} \left(e\right)$

$t = \ln \frac{\frac{1}{2}}{- 0.04 \ln \left(e\right)}$

$t = \ln \frac{\frac{1}{2}}{- 0.04 \left(1\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} t \approx 17.3 \textcolor{w h i t e}{\frac{a}{a}} |}}}$