AOS1 Topic 8: Exponential and Logarithmic Functions

The function f(x)=ax, where aR+{1}, is an exponential function. The shape of the graph depends on whether a>1 or 0<a<1.

Key values are f(1)=1a, f(0)=1 and f(1)=a.

The maximal domain is R and the range is R+.

The x-axis is a horizontal asymptote.

An exponential function with a>1 is strictly increasing, and an exponential function with 0<a<1 is strictly decreasing. In both cases, the function is one-to-one.

Graphing Transformations of f(x)=ax

Translations

If the translation (x,y)(x+h,y+k) is applied to the graph of y=ax, then the image has equation y=axh+k.

The horizontal asymptote of the image has equation y=k.

The range of the image is (k,).

Example:

Sketch the graph and state the range of y=2x1+2

Solution:

Explanation

The graph of y=2x is translated 1 unit in the positive direction of the x-axis and 2 units in the positive direction of the y-axis.

The mapping is (x,y)(x+1,y+2).

Translation of key points:

  • (1,12)(0,52)
  • (0,1)(1,3)
  • (1,2)(2,4)

The range of the function is (2,).

Reflections

If a reflection in the x-axis, given by the mapping (x, y) → (x, −y), is applied to the graph of y=ax, then the image has equation y=ax.

  • The horizontal asymptote of the image has equation y=0.
  • The range of the image is (,0).

If a reflection in the y-axis, given by the mapping (x, y) → (−x, y), is applied to the graph of y=ax, then the image has equation y=ax. This can also be written as y=1ax or y=(1a)x.

  • The horizontal asymptote of the image has equation y=0.
  • The range of the image is (0,).

Example:

Sketch the graph of y=3x

Solution

Explanation

Reflection of y=3x in the x-axis

The graph of y=3x is reflected in the x-axis using the mapping (x,y)(x,y).

Reflection of key points:

  • (1,13)(1,13)
  • (0,1)(0,1)
  • (1,3)(1,3)

Dilations

For k>0, if a dilation of factor k from the x-axis, given by the mapping (x,y)(x,ky), is applied to the graph of y=ax, then the image has equation y=kax.

  • The horizontal asymptote of the image has equation y=0.
  • The range of the image is (0,).

For k>0, if a dilation of factor k from the y-axis, given by the mapping (x,y)(kx,y), is applied to the graph of y=ax, then the image has equation y=ax/k.

  • The horizontal asymptote of the image has equation y=0.
  • The range of the image is (0,).

Example:

Sketch the graphs of the following:

a) y=3×5x

b) y=0.2×8x

Solution:

Explanation:

(a)

The graph of y=5x is dilated by factor 3 from the x-axis.

The mapping is (x,y)(x,3y).

Dilation of Key Points

  • (−1, 1/5) → (−1, 3/5)
  • (0, 1) → (0, 3)
  • (1, 5) → (1, 15)
(b)

The graph of y=8x is dilated by a factor of 15 from the x-axis.

The mapping is (x,y)(x,15y).

Dilation of Key Points:

  • (1,18)(1,140)
  • (0,1)(0,15)
  • (1,8)(1,85)

Note:

Since 9x2=(912)x=3x, the graph of y=9x2 is the same as the graph of y=3x.Similarly, the graph of y=32x is the same as the graph of y=9x.

A translation parallel to the x-axis results in a dilation from the x-axis. For example, if the graph of y=5x is translated 3 units in the positive direction of the x-axis, then the image is the graph of y=5x3, which can be written as y=53×5x. Hence, a translation of 3 units in the positive direction of the x-axis is equivalent to a dilation of factor 53 from the x-axis.

Combinations of Transformations

Combinations of transformations refer to the application of multiple transformations to a function or a geometric object. These transformations can include translations, reflections, dilations, and rotations, among others. By combining these transformations, you can create more complex changes to the object's position, size, and orientation. For example, you can translate a shape, then reflect it across a line, and finally dilate it with respect to a point. These combinations allow for a wide range of modifications to the original object.

The Exponential Function f(x)=ex

In the previous section, we explored the family of exponential functions f(x)=ax, where aR+{1}. One particular member of this family is of great importance in mathematics. This function has the rule f(x)=ex, where e is Euler’s number, named after the eighteenth-century Swiss mathematician Leonhard Euler.

Euler’s number is defined as follows:

Euler’s Number

e=limn(1+1n)n

To see what the value of e might be, we could try large values of n and use a calculator to evaluate (1+1n)n, as shown in the table on the right. As n is taken larger and larger, it can be seen that (1+1n)n approaches a limiting value (≈ 2.71828).

Values of Euler's Number
n (1+1n)n
100 (1.01)100=2.704813
1,000 (1.001)1000=2.716923
10,000 (1.0001)10000=2.718145
100,000 (1.00001)100000=2.718268
1,000,000 (1.000001)1000000=2.718280

Like π, the number e is irrational: e=2.718281828459045

The function f(x)=ex is very important in mathematics.

Exponential Equations

One method for solving exponential equations is to use the one-to-one property of exponential functions:

ax=ayx=y, for aR+{1}

Index Laws

For all positive numbers a and b and all real numbers x and y:

ax×ay=ax+y axay=axy (ax)y=axy (ab)x=axbx (ab)x=axbx ax=1ax 1ax=ax a0=1

Logarithms

Logarithms

Consider the statement 23=8. This may be written in an alternative form: log28=3, which is read as ‘the logarithm of 8 to the base 2 is equal to 3’.

For aR+{1}, the logarithm function with base a is defined as follows:

ax=y is equivalent to logay=x

Note: Since ax is positive, the expression logay is only defined when y is positive.

Further examples:

  • 32=9 is equivalent to log39=2
  • 104=10,000 is equivalent to log1010,000=4
  • a0=1 is equivalent to loga1=0

Inverse Functions

For each base aR+{1}, the exponential function f:RR,f(x)=ax is one-to-one and so has an inverse function.

Let aR+{1}. The inverse of the exponential function f:RR,f(x)=ax is the logarithmic function f1:R+R,f1(x)=logax.

loga(ax)=x for all xR alogax=x for all xR+

Because they are inverse functions, the graphs of y=ax and y=logax are reflections of each other in the line y=x.

The Natural Logarithm

Earlier in the chapter we defined the number e and the important function f(x)=ex. The inverse of this function is f1(x)=logex. Because the logarithm function with base e is known as the natural logarithm, the expression logex is also written as lnx.

The Common Logarithm

The function f(x)=log10x has both historical and practical importance. When logarithms were used as a calculating device, it was often base 10 that was used. By simplifying calculations, logarithms contributed to the advancement of science, and especially of astronomy. In schools, books of tables of logarithms were provided for calculations and this was done up to the 1970s.

Base 10 logarithms are used for scales in science such as the Richter scale, decibels, and pH. You can understand the practicality of base 10 by observing:

log1010=1 log10100=2 log101000=3 log1010,000=4 log100.1=1 log100.01=2 log100.001=3 log100.0001=4

Laws of Logarithms

The index laws are used to establish rules for computations with logarithms.

Law 1: Logarithm of a Product

The logarithm of a product is the sum of their logarithms:

loga(mn)=logam+logan

Proof

Let logam=x and logan=y, where m and n are positive real numbers. Then ax=m and ay=n, and therefore

mn=ax×ay=ax+y (using the first index law)

Hence

loga(mn)=x+y=logam+logan

For example:

log10200+log105=log10(200×5)=log101000=3

Law 2: Logarithm of a Quotient

The logarithm of a quotient is the difference of their logarithms:

loga(mn)=logamlogan

Proof

Let logam=x and logan=y, where m and n are positive real numbers. Then as before, ax=m and ay=n, and therefore

mn=axay=axy (using the second index law)

Hence

loga(mn)=xy=logamlogan

For example:

log232log28=log2(328)=log24=2

Law 3: Logarithm of a Power

loga(mp)=plogam

Proof

Let logam=x. Then ax=m, and therefore

mp=(ax)p=axp (using the third index law)

Hence

loga(mp)=xp=plogam

For example:

log232=log2(25)=5

Law 4: Logarithm of 1m

loga(m1)=logam

Proof

Use logarithm law 3 with p=1.

For example:

loga(12)=loga(21)=loga2

Law 5: Special Values

loga1=0 and logaa=1

Proof

Since a0=1, we have loga1=0.

Since a1=a, we have logaa=1.

Graphing Logarithmic Functions

The graphs of y=ex and its inverse function y=logex are shown on the one set of axes.

For each base aR+{1}, the graph of the logarithmic function f(x)=logax has the following features:

  • Key values are f(1a)=1, f(1)=0, and f(a)=1.
  • The maximal domain is R+ and the range is R.
  • The y-axis is a vertical asymptote.

A logarithmic function with a>1 is strictly increasing, and a logarithmic function with 0<a<1 is strictly decreasing. In both cases, the function is one-to-one.

Graphing Transformations of f(x)=logax

We now look at transformations applied to the graph of f(x)=logax where a>1. We make the following general observations:

  • The graph of y=loga(mxn), where m>0, has a vertical asymptote x=nm and implied domain (nm,). The x-axis intercept is 1+nm.
Example 1

Exponential equations:

Find the value of x for which 52x4=25x+2.

Solution:

52x4=25x+2 Express both sides of the equation as powers with base 5. =(52)x+2=52x+4 Therefore, 2x4=2x+4 Use the fact that 5a=5b implies a=b. 4x=8 x=2

Example 2

Exponential equations

Solve for x

9x=12×3x27

Solution:

We can write the equation as:

(3x)2=12×3x27

Let y=3x. The equation becomes:

y2=12y27

y212y+27=0

(y3)(y9)=0

Therefore, y=3 or y=9

3x=3or3x=32

x=1orx=2

Example 3

Without the aid of a calculator, evaluate the following:

log232

Solution:

Let log232=x.

Then 2x=32.

2x=25

Therefore x=5, giving log232=5.

Example 4

Laws of Logarithms:

Express the following as the logarithm of a single term:

2loge3+loge162loge(65)

Solution:

2loge3+loge162loge(65) =loge(32)+loge16loge((65)2) =loge9+loge16loge(3625) =loge(9×16×2536) =loge10

Example 5

Logarithm equations:

Solve each of the following equations for x

log2(2x1)=4

Solution

log2(2x1)=4

Convert the logarithmic equation to an exponential equation:

2x1=24

Simplify the right side:

2x1=16

Solve for x:

2x=17

x=172

Example 6

Logarithm equations:

Solve each of the following equations for x

loge(x1)+loge(x+2)=loge(6x8)

Solution:

aloge(x1)+loge(x+2)=loge(6x8)

loge((x1)(x+2))=loge(6x8)

(x1)(x+2)=6x8

x2+x2=6x8

x25x+6=0

(x3)(x2)=0

x=3 or x=2

Note:

The solutions must satisfy x1>0, x+2>0 and 6x8>0. Therefore both of these solutions are allowable.

Example 7

Solve each of the following equations for x

loge(2x+1)loge(x1)=4

Solution:

loge(2x+1)loge(x1)=4

Combine the logarithmic terms using the property of logarithms: logamlogan=loga(mn).

loge(2x+1x1)=4

Convert the logarithmic equation to an exponential equation:

2x+1x1=e4

Solve for 2x+1:

2x+1=e4(x1)

Distribute e4:

2x+1=e4xe4

Rearrange to solve for x:

2xe4x=e41

(2e4)x=(e4+1)

x=e4+1e42

Example 8

Graphing transformation of

f(x)=logax

Sketch the graph of y=3loge(2x).

Solution:


Explanation:

This is obtained from the graph of y=logex by a dilation of factor 3 from the x-axis and a dilation of factor 12 from the y-axis.

The mapping is (x,y)(12x,3y).

  • (1,0)(12,0)
  • (e,1)(12e,3)

Exercise &&1&& (&&1&& Question)

How does the shape of the graph of the exponential function f(x)=ax change based on the value of a?

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Exercise &&2&& (&&1&& Question)

What is the equation of the image after applying the translation (x,y)(x+h,y+k) to the graph of y=ax?

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Exercise &&3&& (&&1&& Question)

For \( k > 0 \), if a dilation of factor \( k \) from the x-axis, given by the mapping \( (x, y) \rightarrow (x, ky) \), is applied to the graph of \( y = a^x \), then the image has equation \( y = ka^x \). What will be the range of the image?

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Exercise &&4&& (&&1&& Question)

Which property is used to solve exponential equations?

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Exercise &&5&& (&&1&& Question)

Find the value of \( x \) for which:

\[ 4^x = 256 \]

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Exercise &&6&& (&&1&& Question)

Without the aid of a calculator, evaluate the following:

\[ \log_3 81 \]

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Exercise &&7&& (&&1&& Question)

Solve the following equations for \( x \):

\[ \log_2 x = 5 \]

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Exercise &&8&& (&&1&& Question)

What is the value of \(\log_a 1\) according to the laws of logarithms?

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Exercise &&9&& (&&1&& Question)

Which logarithm law is applied in the equation \(\log_a (m) - \log_a (n) = \log_a \left(\frac{m}{n}\right)\)?

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Exercise &&10&& (&&1&& Question)

Express the following as the logarithm of a single term:

\(2 \log_e 3 + \log_e 16 - 2 \log_e \left(\frac{6}{5}\right)\)

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