Exponential Functions
Exponential functions look somewhat similar to functions you have seen before, in that they involve exponents, but there is a big difference, in that the variable is now the power, rather than the base. Previously, you have dealt with such functions as f(x) = x2, where the variable x was the base and the number 2 was the power. In the case of exponentials, however, you will be dealing with functions such as g(x) = 2x, where the base is the fixed number, and the power is the variable.
Let's look more closely at the function g(x) = 2x. To evaluate this function, we operate as usual, picking values of x, plugging them in, and simplifying for the answers. But to evaluate 2x, we need to remember how exponents work. In particular, we need to remember that negative exponents mean "put the base on the other side of the fraction line".
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So, while positive x-values give us values like these:
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...negative x-values give us values like these:
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You should expect exponentials to look like this. That is, they start small —very small, so small that they're practically indistinguishable from "y = 0", which is the x-axis— and then, once they start growing, they grow faster and faster, so fast that they shoot right up through the top of your graph.
You should also expect that your T-chart will not have many useful plot points. For instance, for x = 4 and x = 5, the y-values were too big, and for just about all the negative x-values, the y-values were too small to see, so you would just draw the line right along the top of the x-axis.
Note also that my axis scales do not match. The scale on the x-axis is much wider than the scale on the y-axis; the scale on the y-axis is compressed, compared with that of the x-axis. You will probably find this technique useful when graphing exponentials, because of the way that they grow so quickly. You will find a few T-chart points, and then, with your knowledge of the general appearance of exponentials, you'll do your graph, with the left-hand portion of the graph usually running right along the x-axis.
Source : http://www.purplemath.com/modules/expofcns.htm
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. For example, the inverse of y = ax is y = logax, which is the same as x = ay.
(Logarithms written without a base are understood to be base 10.)
This definition is explained by knowing how to convert exponential equations to logarithmic form, and logarithmic equations to exponential form. Examples:
1. Problem: Convert to logarithmic form:
8 = 2x
Solution: Remember that the logarithm
is the exponent.
x = log2 8
2. Problem: Convert to exponential form:
y = log3 5
Solution: Remember that the logarithm
is the exponent.
3y = 5
The figure below is a little chart that always helped us remember how to convert from exponential to logarithmic form and from logarithmic to exponential form.
Sometimes you can solve equations containing logarithms by changing everything in logarithmic form to exponential form. Example:
3. Problem: Solve log2 x = -3.
Solution: Convert the logarithm to exponential
form.
2-3 = x
x = (1/8)
There are five special rules that you ought to always have in mind when working with logarithms. They will help you in such tasks as simplifying expressions containing logarithms and solving equations containing logarithms. They are outlined below.
1. For any positive numbers x and y, loga (x * y) = loga x + loga y when a <> 1. Example:
4. Problem: Simplify: log2 x + log2 6.
Solution: log2 (x * 6)
2. For any positive numbers x and p, loga xp = p * loga x. Example:
5. Simplify: logb 9-x
Solution: -x * logb 9
3. For any positive numbers x and y, loga (x/y) = loga x - loga y. Example:
6. Problem: Express as a single logarithm:
loga x - 5loga y
Solution: loga x - loga y5
(Using the 2nd rule.)
Use the third rule in reverse.
loga (x/y5)
4. loga a = 1
5. loga 1 = 0
Source : http://library.thinkquest.org/20991/alg2/log.html
Properties of Logarithms
1. alog (c x d) = alog c + alog d
Example : 2log (8) = 2log (2 x 4) = 2log 2 + 2log 4 = 1 + 2 = 3
2. alog (c : d) = alog c - alog d
Example : 3log (9) = 3log (27 : 3) = 3log 27 - 3log 3 = 3 - 1 = 2
3. alog cd = d x (alog c)
Example : 2log 28 = 8 x (2log 2) = 8 x 1 = 8
4. (alog b)(blog c) = alog c
Example : (2log 65)(65log 8 ) = 2log 8 = 3
5. (alog b) : (alog c) = clog b
Example : (7log 64) : (7log 2) = 2log 64 = 6
Source : http://klikbelajar.com/pelajaran-sekolah/pelajaran-matematika/belajar-matematika-logaritma-dan-pangkat-eksponensial/
Logarithmic and Exponential Equations
Exponential Equations
Some exponential equations can be solved by using the fact that exponential functions are one-to-one. In other words, an exponential function does not take two different values to the same number.
Example 1.
3x = 9
3x = 32
The function f(x) = 3x is one-to-one, so it does not take two different values to 9, so x must equal 2.
x = 2
The equation in example 1 was easy to solve because we could express 9 as a power of 3. However, it is often necessary to use a logarithm when solving an exponential equation.
Example 2.
ex = 20
We are going to use the fact that the natural logarithm is the inverse of the exponential function, so ln ex = x, by logarithmic identity 1. We must take the natural logarithm of both sides of the equation.
ln ex = ln 20
Now the left hand side simplifies to x, and the right hand side is a number. It is approximately 2.9957.
x = 2.9957
Example 3.
5x = 16 We will solve this equation in two different ways.
First Approach: We use the fact that log5 5x = x (logarithmic identity 1 again).
5x = 16
log5 5x = log5 16
x = log5 16
x = ln 16 / ln 5, by the change-of-base formula.
x = 1.7227 (approximately)
Second Approach: We will use the natural logarithm and property 3.
5x = 16 Take the natural logarithm of both sides.
ln 5x = ln 16
x ln 5 = ln 16
x = ln 16 / ln 5
x = 1.7227 (approximately)
We could have used any logarithm with the second approach. The second approach is the one that you see most often.
Example 4.
200 e0.07t = 500
We first isolate the exponential part by dividing both sides of the equation by 200.
e0.07t = 2.5
Now we take the natural logarithm of both sides.
ln e0.07t = ln 2.5
The left hand side simplifies to 0.07t, by logarithmic identity 1.
0.07t = ln 2.5
t = ln (2.5) / 0.07
t = 13.1 (approximately)
Logarithmic Equations
When solving exponential equations we frequently used logarithmic identity 1 because it involves applying a logarithmic function to "undo" the effect of an exponential function. When dealing with logarithmic equations we will use logarithmic identity 2 where an exponential function is applied to "undo" the effect of a logarithmic function.
Example 5.
2 log x = 12
We want to isolate the log x, so we divide both sides by 2.
log x = 6
Since log is the logarithm base 10, we apply the exponential function base 10 to both sides of the equation.
10log x = 106
By logarithmic identity 2, the left hand side simplifies to x.
x = 106 = 1000000
Example 6.
7 + 3 ln x = 15 First isolate ln x.
3 ln x = 8
ln x = 8/3
Now apply the exponential function to both sides.
eln x = e8/3
x = e8/3
This is the exact answer. If you use a calculator to evaluate this expression, you will have an approximation to the answer.
x is approximately equal to 14.39.
Example 7.
ln (x + 4) + ln (x - 2) = ln 7
First we use property 1 of logarithms to combine the terms on the left.
ln (x + 4)(x - 2) = ln 7
Now apply the exponential function to both sides.
eln (x + 4)(x - 2) = eln 7
The logarithmic identity 2 allows us to simplify both sides.
(x + 4)(x - 2) = 7
x2 + 2x - 8 = 7
x2 + 2x - 15 = 0
(x - 3)(x + 5) = 0
x = 3 or x = -5
x = 3 checks, for ln 7 + ln 1 = ln 7.
x = -5 does not check, for when we try to substitute -5 for x in the original equation we are taking the natural logarithm of negative numbers, which is not defined.
So, x = 3 is the only solution.
Source : http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/eandl/equations/equations.html
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