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Applications of Logarithms & Exponentials - PowerPoint PPT Presentation

Applications of Logarithms & Exponentials. We have now reached a stage where trial and error is no longer required!. Ex1. Ex2. Solve lnx = 3.5 to 3 dec places. Solve e x = 14 to 2 dec places. *********. **********. e x = 14 . lnx = 3.5. lne x = ln14 . e lnx = e 3.5. x = e 3.5.

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We have now reached a stage where trial and error is no longer required!

Ex1

Ex2

Solve lnx = 3.5 to 3 dec places.

Solve ex = 14 to 2 dec places.

*********

**********

ex = 14

lnx = 3.5

lnex = ln14

elnx = e3.5

x = e3.5

x = ln14

x = 33.115

x = 2.64

Check ln33.115 = 3.4999….

Check e2.64 = 14.013

Solve 3x = 52 giving answer to 5 dec places.

*********

3x = 52

ln3x = ln52

xln3 = ln52 (law3)‏

x = ln52  ln3

x = 3.59658

Check: 33.59658 =52.0001….

A radioactive substance decays according to the formula

At = A0e–0.55t

where A0 is the initial quantity of the substance and At is the quantity remaining after t hours.

Find the half-life of the substance to the nearest second.

******

For half-life we need At = 1/2A0

A0e–0.55t = 1/2A0

e–0.55t = 1/2

lne–0.55t = ln 1/2

-0.55t = ln1/2

t = ln1/2  (-0.55)‏

t = 1.260…..hrs

( 0.260… X 60 = 15.616..)‏

( 0.616.. X 60 = 36.96..)‏

t = 1hr 15mins 37secs

Ex5

The number of bacteria on an Agar plate grows according to the formula

An = A0e0.75n

where A0 is the initial number of bacteria and An is the number present after n hours.

How long does it take for the number to increase “tenfold” ?

***********

Tenfold is ten times as much !!

For tenfold we need An = 10A0

A0e0.75n = 10A0

e0.75n = 10

lne0.75n = ln10

0.75n = ln10

n = ln10  0.75

n = 3.07..hrs

( 0.07.. X 60 = 4.206…)

n = 3hrs 4.206..mins

(0.206.. X 60 = 12.4..)

n = 3hrs 4mins 12secs

Check: If A0 = 5 then A3.07 = 5 X e (0.75 x 3.07)

= 5 X e 2.3025

= 49.9957….

 50