We say Logarithm of 1,000,000 to the base 10 is 6.

if a^b=c

(read as a raised to the power b equals c)

then, Logarithm of c to the base a is b.

In short, we say log(to the base a) c = b

Common logarithm uses the base 10,

Natural logarithm uses the base 'e'.

Logarithms, discovered by John Napier, makes calculations far easier.

Multiplying two numbers is simply adding their logarithms (to the same base).

For example, logarithm value (to base 10) of a few numbers are given.

log 2 = 0.30103

log 3 = 0.47712

log 7 = 0.8451

log 10 = 1

and so on.]]>

It is whatever "power" (or "exponent") you need to use to get the base to equal the number.

Usually the base is 10 or e (2.71828...) If it is 10 people say "log" or "the base 10 logarithm", if it is "e", then people just say "ln" or "natural logarithm"

So, the base 10 log of 1000 is 3, and the base 10 logarithm of 0.1 is -1 (because 10^-1 = 0.1)

Play around with it on your calculator, and it becomes easy.

]]>I was taking logarithm of 1/2.

Logarithm of 1/2 is a negative number.

(For example, logarithm(base10) of 1/2 is -0.3010 approximately)

When an inequation is mutliplied by a negative number,

the inequality reverses direction.

Thats it!

eg. 5>2

But when this inequation is multiplied by -2,

-10<-4]]>

Oh, and quite a good approach, mathsy. Also, I don't mean to hold anyone back from solving puzzles... it was just an observation of mine.

]]>(Also, some members don't try them straight away to give others a chance)

I do that on the puzzles board, but if someone puts one on the 'Help Me' board then I do it straight away as I assume that they need help.

Wait a minute... when I say assume it changes it to math!

]]>Don't give the answer too soon, let us all have a go at it ...

(Also, some members don't try them straight away to give others a chance)

]]>That's the only difference I know between equations and inequalities, so I thought it might have something to do with that.

*Tries to edit*]]>

Any precise answer?]]>

Step (1) :- 1/4 > 1/8

Step (2) :- (1/2)^2 > (1/2)^3

Taking log on both sides

Step (3) :- 2 log (1/2) > 3 log (1/2)

Cancelling log (1/2) on both sides

Step (4) :- 2 > 3

Obviously, 2 is not greater than 3,

but I started with a correct inequation;

Where did I go wrong???