Question 745760
I would start with properties of logarithms, like
{{{b*log((a))=log((a^b))}}} and {{{log((x))+log((y))=log((xy))}}}
{{{log((A))=log((A[0]))+0.1t*log((0.8))}}}
{{{log((A))=log((A[0]))+log((0.8^(0.1t)))}}}
{{{log((A))=log((A[0]*0.8^(0.1t)))}}}
{{{A=A[0]*0.8^(0.1t)}}}
 
EXTRA:
{{{A}}} as a function of {{{t}}} is an exponential decay function.
The graph looks like this
{{{graph(300,300,-10,90,-15,135,100*0.8^(0.1x))}}}
If you start with a quantity {{{A[0]}}} and lose 20% of what you have every 10 hours, at time {{{t=10}}} hours you will have 80% of {{{A[0]}}} left
{{{A=A[0]*0.8^(0.1*10)=A[0]*0.8^1=A[0]*0.8}}}
At time {{{t=20}}} hours you will have 80% of that 80% left
{{{A=A[0]*0.8^(0.1*20)=A[0]*0.8^2=A[0]*0.64}}} , which is 64% of {{{A[0]}}}
At time {{{t=30}}} hours you will have 80% of 80% of 80% left
{{{A=A[0]*0.8^(0.1*30)=A[0]*0.8^3=A[0]*0.512}}} , which is 51.2% of {{{A[0]}}}