Question 1007142
WITHOUT MEMORIZED FORMULAS:
{{{t}}} days is {{{t/130}}} half-lives,
After that time, the amount of radioactive nuclei would have been halved {{{t/130}}} times.
So, if you started  with an amount {{{A[0]}}} ,
after that time the amount you would have (halved {{{t/130}}} times) is
{{{A=A[0]*(1/2)^"t / 130"}}} .
You could write that as
{{{A/A[0]=(1/2)^"t / 130"}}}<-->{{{A/A[0]=1/2^"t / 130"}}}<-->{{{A[0]/A=2^"t / 130"}}} .
That is what you get just by thinking it through.
It is not a formula that needs to be memorized
(unless you prefer to let other people do the thinking).
So, you could set
{{{A[0]="100%"}}}
{{{A="100%"-"80%"="20%"}}}
and substituting into {{{A[0]/A=2^"t / 130"}}}  you would have
{{{"100%"/"20%"=2^"t / 130"}}}
{{{5=2^"t / 130"}}}
If you take logarithms on both sides of the equal sign, you get an equally valid equation.
You could use logarithms on any base,
but calculators give you logarithms on bases {{{10}}} and {{{e}}} .
So, using base {{{10))) ,
{{{log(5)=log((2^"t / 130"))}}}
{{{log(5)=(t/130)*log((2))}}}
{{{log(5)/log(2)=t/130}}}
{{{t=130*(log(5)/log(2))}}}
{{{t=highlight(302)}}} (rounded to nearest whole number)
 
THE FORMULAS:
The irrational number {{{e}}} is a popular base for calculus reasons.
So, from  {{{A/A[0]=1/2^"t / 130"}}} , using logarithms on base {{{e}}} you get
{{{ln(A/A[0])=ln((1))-ln(2^"t / 130")}}}
{{{ln(A/A[0])=0-(t/130)*ln(2)}}}
From there, you get
{{{A/A[0]=e^"-ln(2)t /130"}}}<-->{{{A=A[0]*e^"-ln(2)t /130"}}} ,or using the approximate value {{{ln(2)=0.693}}} ,
{{{A/A[0]=e^"-0.693 t / 130"}}}<-->{{{A=A[0]*e^"-0.693 t /130"}}} 
Unfortunately, when asked to "show your work",
your teacher may expect to see one of those formulas written out,
just to prove that you are good at memorizing stuff that does not rhyme.