Logarithmic Applications
by Federico S, Josh M, Jake B, and Layeth K
Examples of Situations with Logs
-exponential growth and decay:
-A=P(1+r)^t
-A=P(1-r)^t
-Half-Life:
A=P(1/2)^(t/h)
-Compound interest/growth:
a=P(1+(r/n))^(nt)
-Continuous interest/growth:
-e=2.71
A=Pe^rt
A=final amount
P=original amount
r=rate (as a decimal, not a percentage)
t=time
n=quantity of time in which interest is compounded
h= half-life
Continuous Growth
Formula: A=Pe^(rt)
-ln=log base e
Step One:
29,000=800e^(42r)
36.25=e^(42r)
42r=ln36.25
r=(ln36.25)/42
r=.085
Step Two:
A=Pe^(rt)
A=29000e^(.085*2)
A=$34,373.84
The Studebaker will have a value of $34,373.84 by 2011.
Compound Interest
Josh invests $3000 that he earned by selling Edward, his pet albino snake. If his account gives 2% interest, compounded quarterly, how many years will it take him to earn at least $4000.
Formula:
A=P(1+(r/n))^(nt)
The Solution:
4000=3000(1+(.02/4))^(4t)
4000=3000(1+.005)^(4t)
(4/3)=1.005^(4t)
4t=log (base 1.005) of (4/3)
4t= (log 4/3)/(log 1.005)
t=57.6801/4
t=14.42 (round it up)
It will take 15 years for Josh to earn at least $4000.
Half-Life
Uranium 235 has a half-life of 2700 years. If the original amount of uranium was 60 grams, how many years will it take for 1.875 grams to remain?
The Formula:
A=P(1/2)^(t/h)
The Solution:
1.875=60(1/2)^(t/2700)
(1.875/60)=(1/2)^(t/2700)
(t/2700)=log (base 1/2) of (1.875/60)
t/2700=5
t=13,500
The uranium 235 will take 13,500 years to decrease to 1.875 grams.