The weight of radioactive uranium (grams) remaining after t years is given by the function below where t is greater than or equal to 0. How do you find the time required for the weight to fall to 25% of its original value?

Question

#W##(t)# = #W_0# x #2^-0.0002t#
The t is supposed to be multiplied by -0.0002, Is part of the power.

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Answer 1 · 2018-01-24T09:27:41

See below.

#W(t)=W_0(2)^(-0.0002t)#

We first need to find the mass at the start. This will be at #t=0#

Plugging in #t=0#:

#W(t)=W_0(2)^(-0.0002(0))#

#W(t)=W_0(2)^(0)=>W_0=1#

We need 25% of original value, so we need 25% of 1, i.e.
#0.25# grams

Now, after time #tcolor(white)(88)# #W(t)=0.25# grams:

Using our equation:

#0.25=(1)(2)^(-0.0002t)#

We now solve for #t#

Taking natural logarithms of both sides:

#ln(0.25)=-0.0002ln(2)t#

Dividing by #-0.0002ln(2)#

#ln(0.25)/(-0.0002ln(2))=t=>t=10000# years