© Boardworks Ltd 2006 1 of 23 Contents © Boardworks Ltd 2006 1 of 23 Exponentials and logarithms...

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Exponentials and logarithms

The exponential function

The natural logarithmic function

Equations involving ex and ln x

Examination-style question



Exponential functions

Remember, the general form of an exponential function to the base a is:

f(x) = ax where a > 0 and a ≠1.f(x) = ax where a > 0 and a ≠1.

In both cases the graph passes through (0, 1) and (1, a).

When 0 < a < 1 the graph ofy = ax has the following shape:

y

x

y

x

1 1

When a > 1 the graph of y = ax has the following shape:

(1, a)(1, a)


(1, a)

x

1

0

y = xy = ax

y

1

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.

(a, 1)

y = loga x


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Gradients of exponential functions

Look at the gradient function for some exponential functions:


The number e

This number is denoted by e and is an irrational number.

e = 2.718281828459045235 (to 18 d.p.) e = 2.718281828459045235 (to 18 d.p.)

You can find this number on most scientific calculators by pressing ex and then 1.

(This is not to be confused with an exponential function, which is any expression of the general form ax, where a is a constant).

The function ex is called the exponential function.The function ex is called the exponential function.

For an exponential function f(x) = ax, the value of a for which f(x) = f ’(x) is approximately 2.718.


Transformations of f(x) = ex


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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.

If we start with the equation

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

y

x

y = ex


x0

y = x

1

y = exy

Natural logarithms

A logarithm to the base e is called a natural logarithm.

loge x is written as ln xloge x is written as ln x

So,

We can sketch the graph of y = ln x by reflecting the graph of y = ex in the line x = y.

ln x is the inverse function of exln x is the inverse function of ex

1

y = ln x


Transformations of f(x) = ln x


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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)ln a + ln b = ln (ab)

aa b

bln ln = ln

ln an = n ln a ln an = n ln a

It is also helpful to realise that

ln ex = xln ex = x


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.)



Solve e7x – 4 = 6

Taking natural logarithms on both sides gives

ln (e7x – 4 ) = ln 6

7x – 4 = ln 6

x = 0.827 (to 3 s.f.)

7x = ln 6 + 4

ln6 + 4=

7x



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.)


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 – 75 7

=8

ex

x = 17.7 (to 3 s.f.)


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The function f is defined by

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.

f 1( ) = 3 4xx e x



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, 13 4 = 0xe 13 = 4xe

1 43=xe

1 43ln = lnxe

431= lnx 43= ln 1x = –0.71 (to 2 d.p.)

So, A = (0, 4.15) and B = (–0.71, 0)



c) Let +1= 3 4xy e +1+ 4 = 3 xy e

+1+ 4=

3xy

e

+ 4ln = +1

3

yx

+ 4= ln 1

3

yx

f 1 + 4( ) = ln 1

3

xx

The domain of f –1(x) is x > 4 and the range is f

–1(x) .



d) y = f –1(x) is a reflection of y = f (x) in the line y = x.

y = x

x

y

y = –4

y = f(x)

y = f –1(x)

x = –4

(–0.71, 0)

(0, –0.71)

(0, 4.15)

(4.15, 0)

© Boardworks Ltd 2006 1 of 23 Contents © Boardworks Ltd 2006 1 of 23 Exponentials and logarithms...

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