Base e and Natural Logarithms

Evaluate using the following table

n 10

100

1000

10000

100000

100000

ee

n

n

11

Base e and Natural Logarithms

10—2.5937 100—2.7048 1000—2.7169 10000—2.7181 100000—2.7183 1000000—2.7183 e=2.71828182845904523536028747135266

249775724709369995…

Base e and Natural Logarithms

Natural Base e ln x = loge x The system of natural logarithms has the number

called e as its base. (e is named after the 18th century Swiss mathematician, Leonhard Euler.)  e is the base used in calculus.  It is called the "natural" base because of certain technical considerations.

It is used in continual growth/decay situations

Base e and Natural Logarithms

Inverse Property of Base e and Natural Logarithms e ln x = x and ln ex = x

All rules for log’s apply for ln’s ln 732= 2.5958 ln 0.00614= -5.0929

Base e and Natural Logarithms

A=A0ekt

A = end value, Ao=beginning value

After 500 years, a sample of radium-226 has decayed to 80.4% of its original mass. Find the half-life of radium-226.

0.804 = 1e500k

ln ( 0.804) = ln (e500k)

ln ( 0.804) = 500k ln e ln e=1

ln ( 0.804) = 500k

k= (ln 0.804)/500.

This is the exact solution; evaluate the natural log with a calculator to get the decimal approximation k = -0.000436 .

Base e and Natural Logarithms

Since we now know k, we can write the formula (function) for the amount of radium present at time t as

A=A0 e-0.000436 t

Base e and Natural Logarithms

Now, we can finally find the half-life. We set A=1/2 A0 and solve for t.

(1/2)A0=A0 e -0.000436 t

1/2 = e -0.000436 t

ln(1/2) = ln[e -0.000436 t]. ln (1/2) = -0.000436 t t= ln(1/2) /(-0.000436) t = 1590

Base e and Natural Logarithms

2x=64 solve using ln X ln 2= ln 64 X=ln 64/ln 2 X=6

Base e and Natural Logarithms

POPULATION The equation A = A0e rt describes the growth of the world’s population where A is the population at time t, A0 is the population at t =0, and r is the annual growth rate. How long will take a population of 6.5 billion to increase to 9 billion if the annual growth rate is 2%?

9=6.5e.02t

t=16.3 years

Base e and Natural Logarithms

INTEREST Horatio opens a bank account that pays 2.3% annual interest compounded continuously. He makes an initial deposit of 10,000. What will be the balance of the account in 10 years? Assume that he makes no additional deposits and no withdrawals.

Use A = A0e rt A=10000e .023(10)

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