# Question #6df28

##### 2 Answers

Let

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

Mathematically,

#(A(tau))/(A(0)) = 1/2#

or

#1/2 A(0) = A(tau)# .

Thus, proceed to solve the above equation for

#1/2 xx 450 e^(-0.04 xx 0) = 450 e^(-0.04tau)#

#1/2 xx e^(0) = e^(-0.04tau)#

#1/2 = e^(-0.04tau)#

Take

#ln(1/2) = ln(e^(-0.04tau))#

#-ln(2) = -0.04tau#

#tau = ln(2)/0.04#

#~~ 17.3#

Since you are trying to find the half-life of the substance and you know that

Setting

Thus:

#A(t)=450e^(-0.04t)#

#225=450e^(-0.04t)#

From this point on, we solve for

Start by dividing both sides of the equation by

#color(red)(color(black)(225)/450)=color(red)((color(black)(450e^(-0.04t)))/450)#

#1/2=e^(-0.04t)#

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

#ln(1/2)=ln(e^(-0.04t))#

Using the natural logarithmic property,

#ln(1/2)=-0.04t*ln(e)#

Solving for

#-0.04t=ln(1/2)/ln(e)#

#t=ln(1/2)/(-0.04ln(e))#

#t=ln(1/2)/(-0.04(1))#

#color(green)(|bar(ul(color(white)(a/a)t~~17.3color(white)(a/a)|)))#