## y = b^{x } ≡ x = log_{b} y

So, what is a logarithm? Google defines it as “a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.”

10^{1}=10

10^{2}=100

10^{3}=1000

10^{4}=10000

**A Problem We Are Trying To Solve**

If we had 10^{x} = 5000. What is the value of ^{x}?

Note: On my Casio FX-570W this is log(1) = 0

Log_{10 }1 = 0

Log_{10 }2 = 0.3103

Log_{10 }3 = 0.4771

Log_{10 }4 = 0.6021

Log_{10 }5 = 0.6989

Log_{10 }10 = 1

Log_{10 }100 = 2

Log_{10 }1000 = 3

We can infer from this that** Log _{10 }10^{k} = k**

The following are equivalence (≡) equations. I.e. the same but a different way of writing.

Log_{10 }10 ≡ 10^{1 }

Log_{10 }100 ≡ 10^{2}

Log_{10 }1000 ≡ 10^{3 }

**Exponential Form **

y = b^{x}

**Logarithmic Form**

x = log_{b} y

**Examples**

5000 = 10x (form of y = b^{x} ) ≡ x = Log_{10 }5000 (form of x = log_{b} y) ≡ 3.69897004…

**Covert Exponential Form to Logarithmic Form**

- 2
^{5}= 32 - 3
^{-1}= 1/3 - 4096 = 4
^{6}

My first step would be to convert the Exponential Forms to standard y = b^{x}

- 2
^{5}= 32 ≡**32 = 2**^{5} - 3
^{-1}= 1/3 ≡**0.3 = 3**^{-1} - 4096 = 4
^{6 }≡**4**^{6}= 4096

Therefore,

- 2
^{5}= 32 ≡ 32 = 2^{5 }≡**5 = Log**_{2}32 - 3
^{-1}= 1/3 ≡ 0.3 = 3^{-1}≡**-1 = Log**_{3}0.3 - 4096 = 4
^{6 }≡ 4^{6}= 4096 ≡ 6**= Log**_{4}4096

**Covert Logarithmic Form to Exponential Form **

- Log
_{9}81 = 2 - Log
_{8}4 = 2/3 - -1.5 = Log
_{25}1/125

convert the Logarithmic Forms to standard x = Log_{b} y

- Log
_{9}81 = 2 ≡**2 = log**_{9 }81 - Log
_{8}4 = 2/3 ≡**2/3 = Log**_{8}4 - -1.5 = Log
_{25}1/125**(already in standard form)**

Therefore,

- Log
_{9}81 = 2 ≡ 2 = log_{9 }81 ≡**81 = 9**^{2} - Log
_{8}4 = 2/3 ≡ 2/3 = Log_{8}4 ≡**4 = 8**^{2/3} - -1.5 = Log
_{25}1/125 (already in standard form) ≡**1/125 = 25**^{-1.5}