Lets look at the graph of y = x².

Picture in your mind the graph of y = x². How do you find the inverse of this graph? Geometrically you would reflect the graph over the line y = x. Algebraically you would switch the x & y and then solve for y.
y = x²
x = y²
x^1/2 = (y²)^1/2
x^1/2 = y
Now look at the graph of y = 6^x:
          

First lets find its inverse geometrically:

Now find its inverse algebraically. How do we do this? If we switch the x & y we end up with the y in the exponent. How do we deal with this?You can not multiply or use roots. This is where logarithms com in.

x = b^y is equivalent to y = log_b x. This formula can be used when solving for a variable in the exponent.

There are several types of special logarithms. The first & most frequent are common logarithms. Common logarithms are those whose base (b) is 10. y = 10^x, 1000000 = 10^3x, or y = log₁₀x are all common logarithms. You do not need to specify the base for common logarithms since it is already know to be 10. So 1000000 = 10^3x written in logarithmic form is 3x = log 1000000.

Another popular logarithm deals with the natural number e. e is equal to 2.71828182... or . Logarithms with a base e are called natural logarithms. A natural logarithm is written as y = e^x or x = ln y. ln is used instead of log to represent a logarithmic function with a base of e.

Laws of Logarithm

There are several laws of logarithms that make life solving them a whole lot easier.

log (ab) = log a + log b

Remember when we said if a = 10^m & b = 10ⁿ then ab = 10^(m+n). Converting these equations into logs gives us log a = n, log b = m, and log ab = (n+m). Therefore log ab = log a + log b.

Using the same reasoning we can show that

log (a/b) = log a - log b

&

log aⁿ = n log a

You can sue these properties on not only common logs but on natural or any other type of logarithm as well.

Without a calculator for numbers between 1.0 and 9.9 you can use the log table bellow to determine a logarithm. For any number greater or less than 1.0 to 9.9 you would use log ab and the table below to solve for their exact value.

log 380 = log (3.8 x 10²) = log 3.8 + log 10²
log 3.8 = .580          log 10² = 2 log 10
log 10 = 1     therefore     2 log 10 = 2 x 1 = 2
therefore...
log 380 = log 3.8 + log 10²
log 380 = .580 + 2
log 380 = 2.580

Now what if we wanted to find the number x whose log is 2.670. First we must separate the number at the decimal point. So we now have 2 & .670.

log n = 2.670
log n = 2 + .670
log n = log 10² + .670

Now we must look at the table to see what has its log = to .670. There is no exact value in the table for .670 so we must interpolate. log 4.6 = .663 & log 4.7 = .672 so the answer is between 4.6 & 4.7. . 670 is closer to .672 than .663 by 7/9 so the right answer is approximately 4.68. Put back into the original problem

log n = log 10² + log (4.68)
log n = log (10² x 4.68)
log n = log 468
n = 468

log n = -2.180
log n = -3 + .820
log n = log 10^-3 + log 6.6
log n = log (10^-3 x 6.6)
log n = log 6.6(10^-3)
n = 6.6(10^-3)