Table of Contents

**Definition of Logarithmic Functions**

Logarithmic and exponential function has close relation. It is because logarithmic inverse of exponential function.

Any exponential function can be changed to be logarithmic function and any logarithmic function can be changed to be exponential function.

For example there is exponential function x = a^{y}, so the logarithmic function is ^{a}log x = y or can be written as log_{a} x = y (read it as *log base a to x*).

In another words, it can said that logarithmic function is determining process to get the power of a value.

Another example is 10x10x10x10 = 10^{4} or it is said 10^{4} = 10000 in exponent function. But in logarithmic function, it will be ^{10}log10000 = 4 or can be written log_{10}10000 = 4.

But there is special for base 10. It just write as log10000 = 4.

So, based on the example, logarithmic function is

**If there is a**^{y}** = x is equivalent with y = log**_{a}**x or y = **^{a}**logx**

Note: a is any number such that a>0, a≠1, and x>0.

**Logarithmic Rules**

Remembering the concept of exponent and logarithmic, there are logarithmic rules:

1. Log_{b}b = 1

It’s because b^{1} = b

2. Log^{b}1 = 0

It’s because b^{0} = 1

3. Log_{b}b^{x} = x

It’s because b^{x} = b^{x}

4.

It’s because log_{b}x = log_{b}x

There are also onther rules that based on exponent and logarithmic concept and the operation.

5. Log_{b}(MN) = log_{b}M + log_{b}N (product rule)

6. Log_{b}(M/N) = (log_{b}M)/(log_{b}N) (division rule)

7. Log_{b}M^{x} = x. log_{b}M (power rule)

8. If log_{b}M = log_{b}N then M = N (equality rule)

9. log_{b}M = (log M)/(log b)

**Logarithmic Examples**

Solve the following problems:

- if Log
_{y}8 = 3 then determine the value of*y*!

Based on the concept of logarithmic,

Log_{y}8 = 3 means y^{3}= 8 Þy = 2

2. if Log_{2}(1/8) = y then determine the value of *y*!

Log_{2}(1/8) = y means 2^{y} = 1/8 Þ y = -3 it is because 2^{-3} = 1/8

3. If Log 3 = 0.477 and log 2 = 0.301, then log 18 = …

Log 18 = log (2.9)

= log (2. 3. 3)

= log 2 + log 3 + log 3 (remember the rule)

= 0.301 + 0.477 + 0.477

= 1.255

4. 2 . log_{6}16 – 3 . Log_{6}4 + log_{6}9 = …

2 . log_{6}16 – 3 . Log_{6}4 + log_{6}9 = 2. log_{6}2^{4} – 3. Log_{6}2^{2} + log_{6}3^{2} (remember the rules)

= log_{6}2^{8} – log_{6}2^{6} + log_{6}3^{2}^{}

= log_{6}((2^{8}/2^{6}).3^{2})

= log_{6}(36)

= 2 (because 6^{2} = 36)

5. If log_{3}5 = a and log_{5}7 = b, then determine log_{7}225 = …

First step: determine logarithm that we need to solve the problem.

log_{3}5 = a Þ (log 5)/(log 3) = a Þ log 3 = (1/a). log 5.

Log_{5}7 = b Þ (log 7)/(log 5) = b Þ log 7 =b. log 5.

Second step: use log 7 and log 3 and substitute to the problem

Log_{7}225 = (log 225)/(log 7)

= (log 5^{2}.3^{2})/(log 7)

= (log 5^{2} + log 3^{2})/(log 7)

= (2.log 5 + 2.log 3)/log 7

= (2.log 5 + 2.(1/a)log 5)/b.log 5 (substitute the value from step 1)

= ((2+(2/a)).log 5) /b.log 5 (eliminate log 5)

= (2a+2)/a . (1/b)

=(2a+2)/ab