Prove that loga(m×n) = loga(m) + loga(n)

1. Prove that

loga(m×n) = loga(m) + loga(n)

Proof:

Put loga m = x

Loga n = y

Then by defn

m = a^x and n = a^y

चूकी ( m×n = a^x +a^y = a^ x+y)

Thus by defn we get

x+y = loga(m×n)

Hence

loga(m×n) = loga(m) + loga(n),

2. Prove that

loga(m/n) = loga m - loga n.

Proof: 

Put loga m = x

Loga n = y

Then by defn

We have 

m = a^x and n = a^y

चूकी ( m/n = a^x /a^y = a^ x-y)

So From defn we get

 loga(m/n) = x-y 

Hence

loga(m/n) = loga(m) - loga(n),

3. Prove that 

Loga (m^n) = nloga(m)

Sol: 

Put loga m = x 

Then by defn

m = a^n

m^n = (a^x) + a^nx

Hence

loga (m^n) = nx = n loga m.

4. Prove that

loga m = logb m × loga b.

Sol:

Let x = loga m,

And y = loga b

Then

a^x = m

And

b^y = m

♣ a^x = b^y =⟩ (a^x)¹/y = b

                        =⟩ a^ x/y = b

♣ loga b = x/y = loga m/ loga b

Hence

loga m = logb m × loga b.

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