Using and Deriving
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Exercises for This Topic Index of On-Line Topics Everything for Calculus Everything for Finite Math Everything for Finite Math & Calculus Utility: Function Evaluator & Grapher |
Logarithms
We start by reviewing the basic definitions as in Section 2.3 of Calculus Applied to the Real World. If you like, you can also take a look at the topic summary material on logarithms.
Base b Logarithm
The base b logarithm of x, logbx, is the power to which you need to raise b in order to get x. Symbolically,
1. logbx is only defined if b and x are both positive, and b 1. 2. log10x is called the common logarithm of x, and is sometimes written as log 10. 3. logex is called the natural logarithm of x and is sometimes written as ln x. Examples
Here are some for you to try |
Example 1 Calculating Logarithms by Hand
Algebraic Properties of Logarithms
The following identities hold for any positive a 1 and any positive numbers x and y.
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Example 2 Using the Properties of Logarithms
Let a = log 2, b = log 3, and c = log 5. Write the following in terms of a, b, and c.
Note If any answer you give is not simplified -- for instance, if you say a + a instead of 2a -- it will be marked wrong.
Q Where do the identities come from?
A Roughly speaking, they are restatements in logarithmic form of the laws of exponents.
Q Why is logaxy = logax + logay ?
A Let s = logax and t = logay. In exponential form, these equations say that
Here is an intuitive way of thinking about it: Since logs are exponents, this identity expresses the familiar law that the exponent of a product is the sum of the exponents.
The second logarithmic identity is shown in almost the identical way, and we leave it for you for practice.
Q Why is loga(xr) = r logax ?
A Let t = logax. Writing this in exponential form gives
Identity (d) we will leave for you to do as practice.
Q Why is loga(1/x) = -logax ?
A This follows from identities (b) and (d) (think about it).
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Q Why is
logax | = | log a | = | ln a | ? |
s = | logba |
Since logarithms are exponents, we can use them to solve equations where the unknown is in the exponent.
Example 3 Solving for the Exponent
Solve the following equations for x.
Solution We can solve both of these equations by translating from exponential form to logarithmic form.
(a) Write the given equation in logarithmic form:
4-x2 = 1/64 | Exponential Form | |
log4(1/64) = -x2 | Logarithmic Form | |
Thus, | -x2 = log4(1/64) = -3 | giving | x = 31/2. |
(b) Before converting to logarithmic form, first divide both sides of the equation by 5:
5 (1.12x+3) = 200 | ||
1.12x+3 = 40 | Exponential Form | |
log1.140 = 2x+3 | Logarithmic Form | |
This gives | 2x + 3 = ln 40/ln 1.1 38.7039, | Identity (e) |
so that | x 17.8520. |
You can now either go on and try the exericses in the exercise set for this topic.
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Exercises for This Topic Index of On-Line Topics Everything for Calculus Everything for Finite Math Everything for Finite Math & Calculus Utility: Function Evaluator & Grapher |