# 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=P*e*^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=P*e*^(rt)

-ln=log base *e*

__Step One:__

29,000=800*e*^(42r)

36.25=*e*^(42r)

42r=ln36.25

r=(ln36.25)/42

**r=.085**

__Step Two:__

A=P*e*^(rt)

A=29000*e*^(.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.