Arthur D. answered 07/29/20
Forty Year Educator: Classroom, Summer School, Substitute, Tutor
y=ab^x
24=30(1/2)^(t/28)
24/30=(1/2)^(1/28)
0.8=(0.5)^(t/28)
log(0.8)=log((0.5)^(t/28))
log(0.8)=(t/28)log(0.5)
log0.8/log0.5=t/28
-0.969/-0.30103=t/28
0.32189=t/28
t=28(0.32189)
t=9.013 years
100%-78%=22%
22=78(1/2)^t/30
22/78=(1/2)^t/30
0.2820513=(0.5)^(t/30)
log(0.2820513)=log((0.5)^(t/30))
log(0.2830513)=(t/30)log(0.5)
log(0.2830513)/log(0.5)=t/30
-0.54967189/-0.30103=t/30
1.82597=t/30
t=30*1.82597
t=54.779 seconds