study material-mathematics-algebra-logarithms

# Logarithms - 1

__Definition__:-

We define log as, a^{y} = x than y = log_{a}x. in log_{a}x. both x and a are positive ie. x > 0 and a > 0 and also a 1.

__Dumb Question__:- Why a cannot be 1 ?

__Ans__:- Suppose a is 1 then let us attempt to y such that y = log_{1}x.

Now according to definition of log.

1^{y} = x.

But no matter what power we raise to 1 the answer will be. 1 only so we will never never be able to find y.

Hence a cannot be 1.

__Some important formulae__:- (Formulaes marked with * are important. This is not be printed)

1. log_{a}a = 1.

2. log_{any}1 = 0.

3. log_{c}a = log_{b}a.log_{c}b.

__Why ?__

Let log_{b}a = x and log_{c}b = y

So, by definition, a = b^{x} ………………………………. (i)

b = c^{y} ……………………………….. (ii)

Using (i) & (ii) a = c^{xy}

Now taking log on both sides.

log_{c}a = xy

log_{b}a.log_{c}b.

__Illustration - 1.__

Find value of log_{2}10.log_{10}2 ?

Using formula 3 we get.

log_{2}10.log_{10}2 = log_{10}10

Now using formula 1 we get

log_{10}10 = 1

Hence log_{2}10.log_{10}2 = 1

4. log_{a}(m.n) = log_{a}m + log_{a}n

__Why ?__

Let log_{a}m = x

log_{a}n = y

So, m = a^{x} and n = a^{y} [Using definition]

m.n = a^{x}.a^{y} = a^{x + y}

=> log_{a}(mn) = log_{a}a^{x + y}

= x + y

= log_{a}m + log_{a}n.