Transcript x - MR LEONG

Exponentials and logarithms
Contents
Exponentials and logarithms
The exponential function
The natural logarithmic function
Equations involving ex and ln x
Examination-style question
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Exponential functions
Remember, the general form of an exponential function to the
base a is:
f(x) = ax where a > 0 and a ≠1.
When a > 1 the graph of y = ax
has the following shape:
When 0 < a < 1 the graph of
y = ax has the following shape:
y
y
1
1
(1, a)
x
(1, a)
x
In both cases the graph passes through (0, 1) and (1, a).
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Logarithmic functions
A logarithm is the inverse of an exponential function so that if
f(x) = ax, then f –1(x) = loga x.
The graph of y = loga x is therefore a reflection of y = ax in the
line y = x.
y = ax
y
(1, a)
1
0
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y=x
y = loga x
(a, 1)
1
x
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The exponential function
Contents
Exponentials and logarithms
The exponential function
The natural logarithmic function
Equations involving ex and ln x
Examination-style question
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Gradients of exponential functions
Look at the gradient function for some exponential functions:
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The number e
For an exponential function f(x) = ax, the value of a for which
f(x) = f ’(x) is approximately 2.718.
This number is denoted by e and is an irrational number.
e = 2.718281828459045235 (to 18 d.p.)
You can find this number on most scientific calculators by
pressing ex and then 1.
The function ex is called the exponential function.
(This is not to be confused with an exponential function, which is
any expression of the general form ax, where a is a constant).
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Transformations of f(x) = ex
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The natural logarithmic function
Contents
Exponentials and logarithms
The exponential function
The natural logarithmic function
Equations involving ex and ln x
Examination-style question
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The inverse of the exponential function
Looking at the graph of y = ex we can see that it is a one-to-one
function.
y
If we start with the equation
y = ex
x
y = ex
we can find the inverse by
interchanging the x and the y and
making y the subject of the
formula.
x = ey
Remember that if x = ay then loga x = y. So, we can write this
using logarithms as:
y = loge x
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Natural logarithms
A logarithm to the base e is called a natural logarithm.
loge x is written as ln x
So,
ln x is the inverse function of ex
We can sketch the graph of y = ln x by reflecting the graph of
y = ex in the line x = y.
y
y = ex
y=x
y = ln x
1
0
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x
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Transformations of f(x) = ln x
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Contents
Equations involving ex and ln x
Exponentials and logarithms
The exponential function
The natural logarithmic function
Equations involving ex and ln x
Examination-style question
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Revision of the laws of logarithms
When solving equations involving expressions of the form ln x
and ex you may need to use the laws of logarithms as applied to
natural logarithms:
ln a + ln b = ln (ab)
a
ln a  ln b = ln
b
ln an = n ln a
It is also helpful to realise that
ln ex = x
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Equations of the form eax + b = p
Equations of the form eax + b = p can be solved by taking natural
logarithms on both sides of the equation. For example,
Solve ex = 8
Taking natural logarithms on both sides gives:
ln ex = ln 8
x = ln 8
This can be evaluated using a calculator to give the solution
x = 2.08 (to 3 s.f.)
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Equations of the form eax + b = p
Solve e7x – 4 = 6
Taking natural logarithms on both sides gives
ln (e7x – 4 ) = ln 6
7x – 4 = ln 6
7x = ln 6 + 4
ln6 + 4
x=
7
x = 0.827 (to 3 s.f.)
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Equations of the form eax + b = p
Solve e2x – 3ex = 10
By writing e2x as (ex)2 we can see that this is a quadratic
equation in ex.
(ex)2 – 3ex – 10 = 0
(ex + 2)(ex – 5) = 0
ex = –2 or ex = 5
There is no value of x for which ex ≤ 0 so when ex = –2 there is
no solution.
When ex = 5, x = ln 5
= 1.61 (to 3 s.f.)
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Equations of the form ln (ax + b) = q
Equations of the form ln (ax + b) = q can be solved by rewriting
the equation in the form ax + b = eq. For example,
Solve ln (8x + 7) = 5
This equation can be rewritten in terms of e as:
8x + 7 = e5
8x = e5 – 7
e5  7
x=
8
x = 17.7 (to 3 s.f.)
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Contents
Examination-style question
Exponentials and logarithms
The exponential function
The natural logarithmic function
Equations involving ex and ln x
Examination-style question
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Examination-style question
The function f is defined by
f ( x ) = 3e x  1  4
x
a) Describe the sequence of geometrical transformations by
which the graph of y = 3ex + 1 – 4 can be obtained from that
of y = ex.
b) The graph of y = f(x) crosses the y-axis at point A and the
x-axis at point B. Write down the coordinates of A and B,
working to 2 decimal places.
c) Write an expression for f –1(x) and state its domain and
range.
d) Sketch the graphs of y = f(x) and y = f –1(x) on the same set
of axes and state their geometrical relationship.
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Examination-style question
a) The graph of y = 3ex + 1 – 4 can be obtained from that of y = ex
by stretching it by a scale factor of 3 in the y-direction and
translating it 1 unit left and 4 units down.
b) When x = 0,
y = 3e0 + 1 – 4
= 3e – 4
= 4.15 (to 2 d.p.)
b) When y = 0,
3e x 1  4 = 0
3e x 1 = 4
e x  1 = 34
ln e x 1 = ln 34
x  1= ln 34
x = ln 34  1 = –0.71 (to 2 d.p.)
So, A = (0, 4.15) and B = (–0.71, 0)
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Examination-style question
y = 3e x +1  4
c) Let
y + 4 = 3e x +1

ln 

y+4
= e x +1
3
y+4
= x +1

3 
 y+4
x = ln 
1

 3 
 x+4
 f 1( x ) = ln 
1

 3 
The domain of f –1(x) is x > 4 and the range is f –1(x)  .
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Examination-style question
d) y = f –1(x) is a reflection of y = f (x) in the line y = x.
x = –4
y y = f(x)
y=x
(0, 4.15)
(–0.71, 0)
–1
(4.15, 0) y = f (x)
x
(0, –0.71)
y = –4
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