MATH 11218
WEEK 4
CHAPTERS 8
LOGARITHMS AND EXPONENTIALS
Exponential Expressions:
In an exponential expression a x , a is the base and x is the exponent/power/index.
We will deal with exponential expressions of the form e kx
where e is the exponential constant 2.7183 (to 4dp) and k is a constant.
To simplify exponential expressions we use the index laws as shown in the notes for Week 1- Chapter 5.
Exponential Functions:
An exponential function is of the form y = a bx where a is the base, bx is the exponent and a, b are constants
with a > 0 .
The standard exponential function is y = e x
The graph of the function y = e x approaches the x axis as x → - ∞ , cuts the y axis at (0, 1) and approaches ∞
as x → ∞ (see graph on page 270)
Note:
ex 
1
ex
The graph of y = e –x is a reflection in the y-axis of the graph of y = e x
Applications of Exponential Functions:
If we have exponential growth, the rate of growth is proportional to the amount present at any time.
The formula for the amount present at any time t is N(t) = N0 e kt where N0 is the initial amount.
If we have exponential decay, the rate of growth is also proportional to the amount present at any time but the
formula is N(t) = N0 e - kt where N0 is the initial amount.
With Newton’s Law of Cooling, the rate of loss of temperature is proportional to the difference between the
temperature of the body at ant time, (t), and that of the surrounding environment, e.
The formula for the temperature at any time is
(t) = e + (0 - e) e - kt where 0 is the initial temperature of the body (ie at t = 0)
and k is the positive constant depending on the material of the body
Hyperbolic Functions:
Hyperbolic functions are defined as combinations of exponential functions
hyperbolic cosine
cosh x = ½ ( e x + e –x )
pronounced ‘cosh’
hyperbolic sine
sinh x = ½ ( e x - e –x )
‘shine’
hyperbolic tangent
hyperbolic secant
hyperbolic secant
hyperbolic cotangent
sinh x
e x  e x
 x
cosh x
e  e x
1
sech x 
cosh x
1
x0
cosech x 
sinh x
1
x0
coth x 
tanh x
tanh x 
‘than’
‘sheck’
‘cosheck’
‘coth’
From these hyperbolic functions, we can prove the identities
and
cosh x + sinh x = ex
cosh x - sinh x = e -x
These expressions are used to express exponential expressions as hyperbolic expressions and to prove
other hyperbolic identities.
Reference: Table 1.5 on p278 , Example 1.9 on p276 and Example 1.12 on p 279
To prove an identity, start with the more complicated side, manipulate and simplify it and prove that the
expression can be shown to equal the other (simpler) side.
Note: The equation for a catenary is
y = a cosh( x/b )
as shown on p277
Logarithms:
A logarithm is a power or index.
Definition:
If N = ax
then
x = log a N
ie x is the logarithm of N to base a.
Common logarithms are to base 10 and the base is usually not written. It is understood eg log 1000 = 3
Natural logarithms are to base e and are written as ln
Log Laws:
(ie
log e 5 = ln 5 )
log a ( xy)  log a x  log a y
x
log a    log a x  log a y
 y
log a x n  n log a x
x  a loga x
log a x 
log b x
log b a
(Change of base theorem)
log 1  0
ln ex = x
e ln y = y
Reference: Examples on pp 286 - 288 for simplifying logarithmic expressions
Logarithmic Functions:
The logarithmic function y = logax is the inverse of the exponential function y = a x.
The logarithmic function
y = ln x
is the inverse of the exponential function y = e x.
The graph of the logarithmic function is a reflection in the line y = x of the graph of the exponential function.
Reference: Graph on page 290
Note: The function is only defined for x > 0
Solving Equations with Logarithms and Exponentials:
To solve log equations, we use the definition N = a x  x = log a N
and visa versa.
to change to index equations,
Reference: Examples 3.1 – 3.7 on pp 292 – 295
Applications of Logarithms:
Signal Ratios:
To compare the input vs output levels of an electronic system eg an amplifier, the ratio between th sgnals is
expressed in decibels (dB) where
power gain (dB) =
P 
10 log  o 
 Pi 
where Po is the power of the output signal and Pi is the power of the input signal.
The power of a signal is proportional to the square of its voltage.
 Vo 2 

2 
 Vi 
Voltage gain (dB) = 10 log 

V 
 20 log o 
 Vi 
Reference: pp 297 – 298
Use of log-log and log-linear scales:
If a function is of the form y = abx then taking logs of both sides would yield
log y = log a + x log b
We use a log-linear graph to plot log y against x .
The result is a straight line with gradient = log b and a vertical intercept = log a.
Reference: Examples 4.2 on pp 300
If a function is of the form y = axn then taking logs of both sides would yield
log y = log a + n log x
We use a log-log graph ie we plot log y against log x on log-log paper.
The result is a straight line with gradient = n and a vertical intercept = log a.
Reference: Examples 4.3 – 4.4 on pp 300 – 302