# Logarithmic Applications

## Examples of Situations with Logs

-the pH scale is exponential, with base of ten.

-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

Situation: The 1957 Studebaker Golden Hawk has become a popular car to invest in. In 1967, a Golden Hawk was sold for \$800. 42 years later, in 2009, the same car was worth \$29,000. Use continuous exponential growth to estimate the car's worth in 2011.

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

Situation:

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

The Situation:

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.