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1. The General Exponential Functions 
Since
the natural exponential function e^{x} is differentiable, it's continuous.
Its range is all y > 0. So if b is a positive number, the
IntermediateValue Theorem assures that there exists k such that e^{k} = b.
Thus b^{x} = (e^{k})^{x} = e^{kx}.
Now because e^{k} = b, we
have
k = ln
b. Consequently, b^{x} = e^{(}^{ln}^{ }^{b}^{)}^{x} = e^{x}^{ }^{ln}^{ }^{b.}. Note that the equation b^{x}
= e^{x}^{ }^{ln}^{ }^{b}^{.}
can also be obtained as follows: as e^{x} and ln x are
inverses of each other, we have b = e^{ln}^{ }^{b}, hence b^{x} = (e^{ln}^{ }^{b}^{.})^{x} = e^{x}^{ }^{ln}^{ }^{b}. We define b^{x}
to be e^{x}^{ }^{ln}^{ }^{b}^{.}.
For
any b > 0 and for any x we define b^{x} by: The
function y = b^{x} is called the exponential
function of base b. The general exponential function refers to the exponential function y = b^{x} where the base b can be any positive number. Recall that the natural exponential function y = e^{x} is the exponential function of the
particular base e. 
Recall
that the reason why we require that b > 0 is that we want b^{x} to be defined for all x. If b = 0, b^{x} isn't defined for x = 0
(form 0^{0} isn't defined) or for any x < 0 (no fraction can have denominator 0). If b < 0, b^{x} isn't defined as a real number at for
example x = 1/2 (no negative number has a real
square root). So b^{x} is defined for all x only
if b > 0. Another way to see this
requirement is Eq. [1.1], where we use ln b, requiring that b > 0.
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2. Properties Of The General Exponential Functions 
The
following properties of b^{x} are similar to those of e^{x} as discussed in Section
7.1 Theorem 6.1.
Theorem 2.1 – The Laws Of
Exponents
For
any b > 0 and for any x and y we have: 
Proof
Let's do the proof for part a. Those for the remaining formulas are similar to
it. We have:
b^{x}b^{y} = e^{x}^{ }^{ln}^{ }^{b}e^{y}^{ }^{ln}^{ }^{b} = e^{x}^{ }^{ln}^{ }^{b}^{ + }^{y}^{ }^{ln}^{ }^{b} = e^{(}^{x}^{+}^{y}^{) }^{ln}^{ }^{b} = b^{x}^{+}^{y}.
EOP
the Natural Logarithm Of The General Exponential
In Section
7.2 Theorem 2.1 Part d, we had ln x^{t} = t ln x for
any x > 0 and for any t.
That's the natural logarithm of a power
function. The formula is also true for the natural logarithm of an exponential
function.
Theorem 2.2  Natural
Logarithm Of General Exponential
For
any b > 0 and for any x we have: ln b^{x} = x ln b. 
Proof
ln b^{x}
= ln e^{x}^{
}^{ln}^{ }^{b} = x ln b.
EOP
Solve
3^{x} = 9^{1–}^{x}.
Solution
3^{x} = 9^{1–}^{x},
ln 3^{x} = ln 9^{1–}^{x},
x ln 3 =
(1 – x) ln 9,
x( ln 3 + ln 9) =
ln 9,
EOS
Solving
an equation containing exponential expressions like in this example can be
performed by taking of the natural
logarithm of both sides of the equation.
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3. Differentiation Of The General Exponential Functions 
We
know that (d/dx) e^{x} = e^{x}. In Section
7.1 Introducing e, we got (d/dx) b^{x} = kb^{x}, where e^{k}
= b. Since e^{k} = b, we
have k =
ln b. So (d/dx) b^{x} = (ln
b) b^{x} = b^{x}
ln b.
Theorem 3.1 – The Exponent
Rule
For
any b > 0 and for any x we have: This
differentiation formula is called the exponent rule. 
Proof
Using the chain rule on e^{x}^{ }^{ln}^{ }^{b} we get:
EOP
Remark 3.1
Again
note that (d/dx) b^{x} is not xb^{x}^{–1}!! Keep in mind that b^{x} is an exponential function, not a power
function.
If y = 2^{3}^{x}^{ + 1},
find y'.
y' = 2^{3}^{x}^{ + 1}( ln
2)(3) = 3( ln 2)2^{3}^{x}^{ + 1}.
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4. The General Power Rule 
Now
let's take a look back at the power functions and extend the power rule. We saw
in Section
3.2 Corollary 4.1 that
(d/dx) x^{n} = nx^{n}^{–1}
for all integer n. We saw in Section
3.3 Corollary 4.1 that (d/dx) (u(x))^{r} = r(u(x))^{r}^{–1}(du/dx) for
any
rational number r, from which we have (d/dx) x^{r} = rx^{r}^{–1}
for any rational number r. We now show that (d/dx) x^{a}
= ax^{a}^{–1}
for any real number a, rational or irrational.
Theorem 4.1 – The General
Power Rule
For
any constant a we have: This
differentiation formula is called the general power rule. 
Proof
EOP
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5. Graphs Of General Exponential Functions 
Example 5.1
Sketch
a graph of y = 2^{x}, using information obtained from the
equation.
Solution
Some
More Points: (2, 1/4), (1, 1/2), (1, 2), (2, 4).
The
graph is sketched in Fig. 5.1.

Fig. 5.1 Graph
Of y = 2^{x}. 
EOS
Example 5.2
Sketch
a graph of y = (1/2)^{x}, using information obtained from the
equation.
Solution
Some
More Points: (2, 4), (1, 2), (1, 1/2), (2, 1/4).
The
graph is sketched in Fig. 5.2.

Fig. 5.2 Graph
Of y = (1/2)^{x}. 
EOS
General Case
y = b^{x},
b > 0.
Case 0 < b < 1.
(Think of the graph of y = (1/2)^{x} in Fig. 5.2 where b = 1/2)
Inflection
Points: None.
Case b > 1.
(Think of the graph of y = 2^{x} in Fig. 5.1 where b = 2)
Inflection
Points: None.
The
graphs of y = 2^{x}, y =
(1/2)^{x}, y = 10^{x}, y =
(1/10)^{x}, y = e^{x},
and y = 1^{x} = 1 are sketched in Fig. 5.3. Remark that the graphs
of all exponential functions of the form y = b^{x} go thru the yintercept (0, 1) as b^{0}
= 1 for all b > 0.

Fig. 5.3
Graphs
Of Some Exponential Functions. Graphs
of all exponential functions y = b^{x} go thru

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6. The General Logarithmic Functions 
The
natural logarithm function is defined as the inverse of the natural exponential
function. So it's perfectly natural to define
the general logarithmic function as the inverse of the general exponential
function.
Consider
y = 2^{x}, the exponential function of base 2,
as graphed in Fig. 5.1. Clearly it's onetoone, and
so has an inverse. This
inverse is called the logarithmic function of base 2 or logarithm of base 2, and denoted log _{2}. Thus if y = 2^{x} then x = log _{2} y.
Consequently log _{2} y = exponent on 2 (exponent to which 2
is raised) to get y. The function 1^{x} isn't onetoone and hence has
no inverse. Hence there's no logarithm with base 1 (1^{x} is always 1,
can't be 2 or 10 or any number other than 1). In relation
[6.1] below, we exchange the roles of the letters x and y to conform to the standard presentation of the definition
of an
inverse function.
So log_{b} x =
exponent on b to get x. The
general logarithmic function
refers to the logarithmic function y = log_{b} x,
where the base b can be any real number that's
positive and different from 1. 
Properties

Remarks 6.1
a. In “... the exponent on b to get x”, the quantity appearing after “to
get”, x in
this case, isn't the function (output), it's the
variable (input).
b. The logarithmic function might just as well be called the “ exponent function”. It isn't, because calling it so would
create
confusion with the exponential
function.
The Natural Logarithm Is The
Logarithm Of Base e
By Definition
6.1, the inverse of the natural exponential function e^{x} of base e is
the logarithm function log_{e} of base e. We know
that this inverse is the natural logarithm function ln. So the natural logarithm function is the logarithm function
of the particular
base e.
The
natural logarithm function is the logarithm function of base e: ln x = log_{e} x. 
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7. Properties Of The General Logarithmic Functions 
The
following properties of the general logarithmic functions except the last one
are similar to the corresponding properties of
the natural logarithm function as discussed in Section
7.2 Theorem 2.1. One
of the properties of the general logarithmic
functions that we're going to state and prove is log_{b} xy = log_{b} x + log_{b} y. This relation expresses the logarithm of
the product
xy
of 2 numbers x and y in terms of the
logarithms of the original numbers x and y. This property says
that the logarithm of
the product of 2 numbers equals the sum of the logarithms of the original
numbers.
Theorem 7.1  Properties Of
General Logarithmic Functions

a. Let y = log_{b} 1. Then b^{y}
= 1 = b^{0}.
So log_{b}
1 = y = 0. (Intuitively, the exponent on b to get 1 is 0.)
b. Let y = log_{b} b.
Then b^{y} = b = b^{1}. So log_{b} b = y = 1. (Intuitively, the exponent on b to get b is 1.)
c. Let p = log_{b} x and q = log_{b} y, so that x = b^{p} and y = b^{q}.
Then log_{b} xy = log_{b} b^{p}b^{q} = log_{b} b^{p}^{+}^{q} = p + q = log_{b} x + log_{b} y.
d. log_{b} x + log_{b} (1/x) = log_{b} (x(1/x)) (by part c) = log_{b} 1 = 0. So log_{b} (1/x) = 
log_{b}
x.
e. log_{b} (x/y) = log_{b} (x(1/y)) = log_{b} x + log_{b} (1/y) (by
part c) = log_{b} x  log_{b} y (by
part d).
f. Note that if
t is a positive integer then, by
repetitive applications of part c:
For t = 0, we have log_{b} x^{0} = log_{b} 1 = 0 = 0 log_{b} x.
For any
non0 real number t. Let y = log_{b} x^{t}, so that x^{t} = b^{y}. Then x = b^{y}^{/}^{t}, then y/t = log_{b} x, then y = t log_{b} x. Thus
log_{b} x^{t} = t log_{b} x.
Consequently, for any real number t we have log_{b} x^{t} = t log_{b} x.
g. Let y = log_{b} x. Then x = b^{y}. So log_{c} x = log_{c} b^{y} = y log_{c} b (2nd equation is by part f). Thus
log_{b} x = y = (log_{c} x)/( log_{c} b).
EOP
Corollary 7.1  Logarithms Of
Extended Products And Quotients
For
any real p_{i}, any real q_{j}, any positive integer m, and any positive integer n we have: 
Proof
EOP
EOS
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8. Differentiation Of The General Logarithmic Functions 

Proof
EOP
Derivatives Of Exponentials And Logarithms
For
help in the distinction and memorization of the derivatives of the exponential
and logarithmic functions, we gather them
together in the following table.
Derivatives Of Exponentials And Logarithms

Differentiate:
EOS
In Solution
2 we first simplify the function expression before differentiating. This
simplifies the differentiation greatly.
Solution 2 is simpler than Solution 1 and for this reason is preferable.
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9. Graphs Of The General Logarithmic Functions 
Graphs Of Inverses

Fig. 9.1 b > 1, graph
of y = log_{b} x is
reflection of that of its inverse y = b^{x} about line y = x. 

Fig. 9.2 0
< b < 1, graph
of y = log_{b} x is
reflection of that of its inverse y = b^{x} about line y = x. 
Graph Of y = ln x

Fig. 9.3 Graph
Of y = ln x. 
Example 9.1
Sketch
a graph of y = log_{2}
x, using information from the equation.
Solution
Inflection
Points. Since there's no change of concavity, there's no inflection point.
Some More Points: (1/4, 2), (1/2, 1), (4, 2), (8, 3). ( y = exponent on 2 to get x, so choose xvalues that are integer powers
of 2 such that it's easy to
determine y
mentally.)
The
graph of y = log_{2} x is sketched in Fig. 9.4.

Fig. 9.4 Graph
Of y = log_{2}
x. 
EOS
Example 9.2
Sketch
a graph of y = log_{1/2}
x, using information from the equation.
Solution
Inflection
Points. Since there's no change of concavity, there's no inflection point.
Some More Points: (1/4, 2), (1/2, 1), (2, 1), (4, 2), (8, 3).
The graph
of y = log_{1/2} x is sketched in Fig. 9.5.

Fig. 9.5 Graph
Of y
= log_{1/2} x. 
EOS
General Case
Inflection Points.
Since there's no change of concavity, there's no inflection point.
Case b > 1. (Think of the graph of y = ln x in Fig. 9.3 where b = e > 1 or the graph of y = log_{2} x in Fig. 9.4 where b = 2)
Inflection
Points. Since there's no change of concavity, there's no inflection point.
Graphs
The
graphs of some logarithmic functions are sketched in Fig. 9.6. Remark that the
graphs of all the basic logarithmic
functions go thru the xintercept (1, 0) ( y = log_{2} x is a basic logarithmic function, while y = 3 log_{2} x  4 isn't; it's a
transformed logarithmic function). A way to help memorize the shape of the
graph is the graphs of y = ln
x
= log_{e} x and
y
= log_{1/2} x: since e > 1, the graph
of y
= log_{b} x where b > 1 has a
similar shape as that of y = ln
x;
since 0 < 1/2 < 1,
the graph of y = log_{b} x where 0 < b < 1 has a
similar shape as that of y = log_{1/2} x. So memorize the
graphs of y
= ln x
and y
= log_{1/2} x.

Fig.
9.6 Graphs
Of Some Logarithmic Functions. Graphs of all 
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10. The Common Logarithm 
The
logarithm of base 10, log_{10} x, is called the common logarithm,
and usually is denoted log x,
without writing the base 10.
It's called so because it's commonly used in applications of mathematics. The
exponential of base 10, 10^{x}, which is the inverse
of the common logarithm, is similarly called the common antilogarithm.
Consider
the common logarithm y
= log x. As shown in Fig. 10.1, when x increases for example from 1 to
1,000,000,
log x increases only from 0 to 6. So log x increases slowly. Also see the graph of log_{10}
x in Fig. 9.5. When x increases
from 1 to 10 thus by 10 – 1 = 9 units, log x increases from 0 to 1 hence by 1 – 0 = 1 unit; when x increases from 10 to
100 thus by 100 – 10 = 90 units, log x
increases from 1 to 2 hence by 2 – 1 = 1 unit, still by 1 unit; when x increases from
100,000 to 1,000,000 thus by 1,000,000 – 100,000 = 900,000 units, log x increases from 5 to 6 hence by 6 – 5 =
1 unit, still by
1 unit. The larger x is, the slower log x increases. This can also be seen by
examining the graph of log_{10} x in Fig. 9.5
and the
derivative (rate of change) 1/(x ln 10)
of log x.
Clearly the larger x > 0 is, the smaller (down towards
0) 1/(x ln 10) becomes.
As x > 0 increases, log x increases slowly and at a decreasing rate. (The function y = (0.000,001)x increases slowly but at a
constant rate of dy/dx =
0.000,001;
its graph is a straight line; that of y = log
x is concave down).
Fig. 10.1
Common
logarithm 
Problems & Solutions 
1.
Simplify the following expressions.
Solution
2.
Solve the equation 5^{2}^{x}3^{2}^{x}^{–1}
= 15^{x}^{–2} for x.
Solution
5^{2}^{x}3^{2}^{x}^{–1} = 15^{x}^{–2},
ln 5^{2}^{x}3^{2}^{x}^{–1} = ln 15^{x}^{–2},
ln 5^{2}^{x} + ln 3^{2}^{x}^{–1} = (x – 2) ln 15,
2x ln 5 + (2x – 1)
ln 3 = x ln 15 – 2 ln 15,
x(2 ln 5 +
2 ln 3 – ln 15)
= – 2 ln 15 + ln 3,
3.
Differentiate the following functions.
a.
f (x) =
t^{x}
– x^{t}.
b. g(t) = t^{x} – x^{t}.
c. y = log_{a}_{ }(bx
+ c).
Solution
a.
f '(x) = t^{x} ln t – tx^{t}^{–1}.
b.
g'(t) = xt^{x}^{–1} – x^{t} ln x.
4.
Consider the function y
= (2x)^{3}^{x}.
The variable appears in both the base and the exponent. So the power rule and
exponent
rule of differentiation don't apply
to it directly.
a. Differentiate it by using the
definition of the general exponential function.
b. Differentiate it by first taking
the natural logarithm of both sides and then using implicit differentiation.
c. Are the answers in parts a and b
the same?
Solution
a. y = (2x)^{3}^{x}
= e^{3}^{x}^{ }^{ln}^{ 2}^{x},
y '
= e^{3}^{x}^{ }^{ln}^{ 2}^{x}
(3 ln 2x
+ 3x(2/2x)) = (2x)^{3}^{x}
(3 ln 2x
+ 3) = 3(2x)^{3}^{x}
(ln 2x + 1).
b.
ln y
= ln (2x)^{3}^{x}
= 3x ln
2x,
differentiate first and third sides
implicitly:
y '/y = 3 ln
2x + 3x(2/2x) = 3 ln
2x + 3 = 3( ln
2x + 1),
y '
= y(3( ln
2x + 1)) = 3(2x)^{3}^{x}
(ln 2x
+ 1).
c.
Yes.
Note
Of course in practice just use one method,
unless you're asked to use both methods.
5.
Prove that for any a > 0, any b > 0, and any c > 0, log_{a} b = log_{a} c log_{c} b.
Solution
log_{a} b = log_{a} c log_{c} b.
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