Logarithms
Logarithms
|Definition |The logarithm of a number to a particular base is the power (or index) to which that base must be raised to obtain the number. |
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| |This means that a logarithm is an index. |
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| |The number 8 written in index form is 8 = 23 |
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| |The equation can be rewritten in |
| |logarithm form as [pic] |
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| |The logarithm statement reads |
| | the logarithm of 8, to the base 2 is 3 |
| |or 3 is the logarithm of 8 to the base 2 |
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| |and is equivalent to the index statement |
| |8 equals 2 to the power 3 |
| |or 2 to the power 3 equals 8. |
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| |In general |
| |[pic] [pic] [pic] |
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|Examples |1. Change the following from index form to logarithm form. |
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| |(a) [pic] |
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| |the base is 10, the index is 3, the number is 1000 |
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| |so [pic] |
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| |(b) [pic] |
| |the base is 2, the index is -5, the number is [pic] |
| |so [pic] |
| |(c) [pic] |
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| |the base is 16, the index is[pic], the number is 4 |
| |so [pic] |
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| |2. Change the following from logarithm form to index form . |
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| |(a) [pic] |
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| |the base is 2, the number is 16, the index is 4 |
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| |so [pic] (follow the arrows and read [pic]) |
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| |(b) [pic] |
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| |the base is 10, the number is 10, the index is 1 |
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| |so [pic]0 |
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| |(c) [pic] |
| |the base is 3, the number is[pic] , the index is -4 |
| |[pic] |
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| |In general |
| |[pic] [pic] (follow the arrows and read [pic] ) |
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|Exercise 1 |Write in logarithm form |
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| |[pic] |
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| |Write in index form |
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| |[pic] |
|Evaluating |One method for evaluating logarithms is: |
|logarithms |let x be the value of the logarithm |
| |rewrite in index form |
| |3. solve the indicial equation. |
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|Example 1 | Evaluate [pic] |
| |[pic] = x write x as the value of the logarithm |
| |[pic] |
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| |Therefore [pic] = 3 |
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| |Another method is |
| |write the number in index form, with same base as the logarithm |
| |use the definition of a logarithm to evaluate. |
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| |Evaluate [pic] |
| |125 = 53 125 in index form with base 5 |
| |[pic] equivalent logarithm statement. |
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| |Evaluate [pic] |
| |0.0001 = 10-4 |
|Example 2 |[pic] |
|Exercise 2 |Evaluate the following |
| |(a) [pic] |(b) [pic] |
| |(c) [pic] |(d) [pic] |
| | |(f) [pic] |
| |(e) [pic] | |
| | |(h) [pic] |
| |(g) [pic] | |
|Logarithm laws |The logarithm laws are obtained from the index laws and are: |
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| |[pic] |
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| |Note: It is not possible to have the logarithm of a negative number. |
| |All logarithms must have the same base. |
|Simplifying |The logarithm laws can be used to simplify logarithmic expressions. |
|logarithms | |
|Example 1 |Express the following as a single logarithm. |
| |(Remember the logarithms must have the same base if they are to be added |
| |or subtracted). |
| | (1) [pic] Use the laws for adding and subtracting logarithms. |
| |[pic] |
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| |(2) [pic] [pic] and [pic] must be written as [pic] and [pic] before using addition and subtraction laws. |
| |[pic] |
| |=[pic] |
| |= [pic] |
| |= [pic] |
|Example 2 |Simplify and evaluate |
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| |[pic] |
| |=[pic] write each term in the form [pic] |
| |[pic] use laws for adding and subtracting logarithms |
| |[pic] |
|Exercise 3 |Express as a single logarithm and evaluate, if possible, without using a calculator . |
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| |(a) [pic] |(b) [pic] |
| |(c) [pic] |(d) [pic] |
| |(e) [pic] |(f) [pic] |
|Solving equations |Equations of the type [pic] can be solved for x by rewriting the equation in index form. |
|Example | Solve the following for x. |
| | [pic] |
| |[pic] |
| |Therefore x = 32 |
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| |Logarithms can also be used to solve indicial equations of the form [pic] |
| |We solve these equations by taking logarithms of both sides. By taking the logarithms to the base 10 or the base e (Euler’s |
| |number), we can use a calculator to evaluate logarithms. Logarithms to the base e are often called natural logarithms. |
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| |On the calculator use the log button to evaluate logarithms to the base 10 and the ln button to evaluate logarithms to the base |
| |e. |
|Example |Solve for x giving your answer to three decimal places. |
| | [pic] |
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| |Take logarithms of both sides |
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| |[pic] |
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|Exercise 4 |Solve for the unknown, giving your answer to three decimal places. |
| | |(b) [pic] |
| |(a) [pic] | |
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| |(c) [pic] |(d) [pic] |
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|Exponential growth |Many natural processes exhibit exponential growth or decay. Some examples are growth of bacteria, radioactive decay, discharge of|
|and decay |a capacitor and rate of temperature change. |
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|Example | The number of bacteria present in a sample is given by |
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| |[pic], where t is in seconds. |
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| |Find: |
| | (a) the initial number of bacteria |
| | (b) the time when the number of bacteria reaches 10 000 |
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| |(a) The initial number of bacteria occurs when t = 0 |
| |Substitute t = 0 in the equation for N. |[pic] |
| | |[pic] |
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| | |The initial number of bacteria is 800. |
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| |(b) The number of bacteria, N, is equal to 10 000. |
| |Substitute N = 10 000 in the equation for N. |[pic] |
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| | |[pic] |
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| | |It takes 12.6 sec. For the number of bacteria to reach 10 000. |
| |Take the logarithm to the base e of both sides. | |
|Exercise 5 |The decay rate for a radio-active element is given by |
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| |[pic] |
| |where R is the decay rate in counts/s at time t(s). |
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| |The half–life is the time when R has been reduced to half the initial |
| |value. |
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| |If the initial decay rate is 400 counts/s find: |
| | (a) the decay rate after 1 minute |
| | (b) the half-life of the element. |
| | (c) the time when the decay rate falls to 2 counts/s |
|Exercise 6 |The voltage in a circuit is given by |
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| |[pic] |
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| |If E = 150V, R = 103 ohm, and C =[pic] farad, find |
| |The time (seconds) when V = 73V. |
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|Exercise 7 |The charge Q units on a plate of a condenser t seconds after it |
| |starts to discharge is given by |
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| |[pic] |
| | If Q = 1840 when t = 0.5 and Q = 667 when t = 1 find: |
| |(a ) the value of k |
| | (b) the initial charge Q0 |
| | (c) the time taken for the charge to fall to 1000 units. |
|Graphs of |All logarithm graphs have the same basic shape. |
|logarithmic |Consider the graph of [pic]. |
|functions | |
| |x and y intercepts |
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| |when y = 0 x = 1 i.e the graph cuts the x-axis at the point x = 1. |
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| |when x = 0 y is not defined. However,[pic]therefore |
| |. the graph has a vertical asymptote at x = 0. |
| |behaviour at [pic] |
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| |[pic] |
| |It is not possible to have the logarithm of a negative number, so does not exist at [pic] |
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| |turning points There are no turning points. |
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| |when [pic] Therefore the graph passes through the point [pic] |
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|Answers | | | | |
|Exercise 1 |(a) [pic] |(b) [pic] |(c) [pic] |(d) [pic] |
| |(e) [pic] | |(g) [pic] |(h) [pic] |
| | |(f) [pic] | | |
| |(i) [pic] |(j) [pic] |(k) [pic] |(l) [pic] |
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|Exercise 2 |(a) 2 |(b) [pic] |(c) 7 |(d) 5 |
| |(e) 0 |(f) ( 3 |(g) ( 3 |(h) [pic] |
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|Exercise 3 |(a) [pic] | |(c) [pic] |(d) [pic] |
| | |(b) [pic] [pic] | | |
| |(e) [pic] | | | |
| | |(f) [pic] | | |
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|Exercise 4 |(a) x = 0.699 |(b) x = 11.62 |(c) x = 0.185 |(d) x = ( 1.807 |
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|Exercise 5 |(a) R = 66 counts/s |(b) t = 23 sec. |(c) t = 177 sec. | |
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|Exercise 6 | t =1 sec. | | | |
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|Exercise 7 |(a) k = 0.8814 |(b) [pic] |(c) t = 0.8 sec. | |
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base
y
x
1
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(a,1)
[pic]
base
Index or logarithm
Index or logarithm
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