Working with Exponentials and Logarithms

First, the graphical representation of the most important exponential curves (see math script as well)
Exponential function
 
Exponential function - 1/x
The typical curves everybody should know. The blue curve in the first quadrant (positive x values) corresponds to the energy dependence of the ubiquitous Boltzmann factor exp – (E/kT) Slightly more tricky. Note that the purple branch in the 1. quadrant corresponds to the temperature dependence of the ubiquitous Boltzmann factor exp – (E /kT)
The inverted functions, e.g. y = ln x are easily pictured, too; below the y = ln x and the y = ln (1/x) functions are shown.
Ln functions
The graphs, in case you forgot, also illustrate some basic algebraic relations, e.g.
  • e– x = 1/ex
  • ln x = – ln (1/x)
The essential identities are
ex  =  1
e– x 		  (ex)y  =  ex · y
         
ex · ey  =  ex + y   (ex)1/y  =  ex /y  
             
ex
ey 		  =  ex – y   eln x  =  x 
 
ln (x · y)  =  ln x  +  ln y
     
ln x
y 		 =  ln x  –  ln y
     
ln xy  =  y · ln x
Here some approximations.
ex  =  1  +  x
1! 		 +   x2
2! 		 +  x3
3! 		 + .. 
 
ln (1 + x)  =  x  –   x2
2 		 +  x3
3 		 –   x4
4 		 + ..    

ln (1 – x)  =   –   æ
è 		x  +  x2
2 		 +  x3
3 		 +   x4
4 		 + ..   ö
ø
While many equations contain exponential terms of some variable which "disappear" if you substitute the ln of the variable (as in the Arrhenius plot), we mostly prefer the lg of some observable quantity to the ln. As an example, plotting the vacancy concentration cV in an Arrhenius plot would be straight forward with the ln:
On order to get a straight slope we have to switch to new variables according to
cV  = A · exp –   HF
kT

ln cV   = ln A   –   HF
kT 		 ·  1
T

y  = ln A   –   1
T		  ·  x

for
y  = ln cV  x  =  1
T
For y and x we get straight line with slope –HF/k and intercept = A
What do we have to do if we want to plot lg (cV) instead of ln (cV)?
We have to multiply everything with lg e = 0,4342.., because
lg x  =  (lg e) · (ln x) 1) 
We obtain
lg (cV)  =  0,434 · ln A  –  0,434 ·   HV
k		  ·   1 
T
 
 
1) If this puzzles you, consider
We postulate the equality
lg x  =  M · ln x
and want to find a value for M. Raising everything to the power of 10 gives
10lg x  =  x  =  10M · ln x
This equation can only be fulfilled if
10M  =  e
because then we have eln x = x as required, giving M = lg e
 

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gehe zu 5.3.3 Gleichgewichtskonzentration von atomaren Fehlstellen in Kristallen

gehe zu 5.3.2 Definition der Entropie und erste Anwendung

gehe zu 5.3.4 Darstellungen der Konzentrationsfunktion

gehe zu Lösung Übung 6.3-1

gehe zu Lösungen zur Übung 2.2-5

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