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Introduction to Logarithms
Log questions ask which power a base must be raised to in order to equal a particular given number. Consider this question.$$\log_2{(4)}\;=?$$ The above log question is asking what power of 2 gives the answer of 4, or how many times do you multiply 2 by itself to get 4. The answer of 4 is found by 2 x 2, which is counted as two times. so the answer to the log is 2, as shown here,
$$\log_2{(4)}\;=\;2$$
Another question is, $$\log_3{(27)}\;=?$$
Once again the question is asking how many times do you multiply 3 by itself to get 27 (the number that is known as the argument). Since 3 x 3 x 3 = 27, the answer is 3 because we stated the number three times in the multiplication so therefore,
$$\log_3{(27)}\;=3$$
The logarithmic function, y = logb(x) is the inverse of the exponential function x = by.
The base of the logarithm is b, which is a positive real number. The argument is x, and is a positive value.
Another Logarithm Example
As stated, logarithm questions are asking you how many times you have to multiply a number by itself in order to get a particular number. Log base 2 of the argument 8 is written,
$$\log_2{(8)}\;=$$
The above logarithm asks: how many factors of 2 multiply together to make 8? Since 2 × 2 × 2 = 8, there are 3 factors of 2, so the answer is 3. Another way of saying this is: what power would 2 have to be raised to in order to equal 8? Since you multiply 2 three times, the answer is 3.
$$2\;×\;2\;×\;2\;=\;8,\; $$$$so\; 2^3\;=\; 8,\;$$
$$and\;therefore\;\log_2{(8)} \;= \;3$$
Try answering these similar logarithm questions yourself.
$$a)\;5 $$ $$b)\;2$$ $$c)\;4$$ $$d)\;6 $$
Show Answer
$$c)\;4 $$
$$a)\;3 $$ $$b)\;4$$ $$c)\;2$$ $$d)\;5 $$
Show Answer
$$b)\;4 $$
$$a)\;2 $$ $$b)\;6$$ $$c)\;3$$ $$d)\;4 $$
Show Answer
$$a)\;2 $$
$$a)\;2 $$ $$b)\;6$$ $$c)\;4$$ $$d)\;5 $$
Show Answer
$$b)\;6 $$
$$a)\;2 $$ $$b)\;4$$ $$c)\;3$$ $$d)\;6 $$
Show Answer
$$c)\;3 $$
Logarithms – Notable Cases
Whenever your log argument is (1) the result is 0, for instance,
$$\log_3{(1)}\;=\;0$$
Whenever your log is the same number as the base the answer will be 1, for instance,
$$\log_9{(9)}\;=\;1$$
If there is no base, the base is always assumed to be 10, for instance,
$$\log \; (1000)\;=\;3, \;$$because$$10\; ×\; 10\; ×\; 10 \;=\;1000\; $$or$$10^3\; = \;1000$$
When your log is a number such as .00001 results are negative exponent values, for instance,
$$\log_{10}{(0.00001)}\;=\;-5,\;$$because$$\;\frac{1}{10}\; ×\; \frac{1}{10}\; ×\; \frac{1}{10}\; ×\; \frac{1}{10}\; ×\; \frac{1}{10}\; $$or $$\;10^{-5}\; = \;0.00001$$
In a question like the following, where the fours are the same, the two fours cancel each other out, and the answer is simply 21, so,
$$4^{log_4{21}}\;=\; 21,$$and another example is,$$5^{log_5{48}}= 48$$
Compare this,
$$\log_4{(16)}\;=\;2,\;$$and this, $$\log_4{(\frac{1}{16})}\;=\;-2$$
and compare this,
$$\log_4{(16)}\;=\;2,\;$$and this,$$\log_{16}{(4)}\;=\;\frac{1}{2}$$
We calculated the last result by flipping the fraction, just as the base and argument were flipped.
$$\frac{2}{1}, $$the answer to the first question was flipped to get, $$\frac{1}{2}$$
In cases where the base number is larger than the argument number, for instance, in the above case where you are looking to find the number of times you multiply 16 to get 4, you will always have a fraction as the answer.
Also consider this example:
$$\log_9{(\frac{1}{3})}\;=\;-\frac{1}{2}$$
Whenever you are finding the log of a fraction, the answer will be negative as in the above case.
More examples:
$$\log_2{(16)}\;=\;4$$and$$\; \log_{16}{(2)}\;=\;\frac{1}{4}$$
$$\log_2{(\frac{1}{16})}\;=\;-4$$and$$\log_{16}{(\frac{1}{2})}\;=\;-\frac{1}{4}$$
See if you can answer these five logarithm questions.
$$a)\;1 $$ $$b)\;3$$ $$c)\;2$$ $$d)\;6 $$
Show Answer
$$c)\;2 $$
$$a)\;5$$ $$b)\;6$$ $$c)\;4$$ $$d)\;2 $$
Show Answer
$$a)\;5 $$
$$a)\;\frac{1}{5} $$ $$b)\;\frac{1}{4} $$ $$c)\;-\frac{1}{2} $$ $$d)\;-\frac{1}{5} $$
Show Answer
$$d)-\frac{1}{5} $$
$$a)\;3 $$ $$b)\;2$$ $$c)\;4$$ $$d)\;1 $$
Show Answer
$$b)\;2 $$
$$a)\;6 $$ $$b)\;-4$$ $$c)\;2$$ $$d)\;-5 $$
Show Answer
$$b)\;-4 $$
Logarithms were widely used from the 17th century onwards in many fields, including in engineering, astronomy and navigation. Their introduction is attributed to the independent work and publications of John Napier and Joost Burgi, who were both seeking to simplify mathematical calculations through the application of logarithmic principles. The tables, and slide rules which came later, included logarithms, and made it possible for high accuracy computations to be made quickly and easily. Logarithms enabled simpler addition processes, and replaced complex and time-consuming multiplication. By virtue of the products rule, $$\log_b{(x)}\;+\; \log_b{(y)}\;=\;\log_b{(xy)}$$ the product of two numbers could be found easily through addition, as long as b, x and y were all positive.