Question

prove that log (M×N)=logM+logN
please answer me fast​

Answer 1

Answer:

MARK AS BRAINLIEST

Step 1:

Let m = loga x and n = loga y

Step 2: Write in exponent form

x = am and y = an

Step 3: Multiply x and y

x • y = am • an = am+n

Step 4: Take log a of both sides and evaluate

log a xy = log a am+n

log a xy = (m + n) log a a

log a xy = m + n

log a xy = loga x + loga y

Answer 2

Question :

Proof that log(m×n) =log m +log n

Logarithm function :

The Logarithm function is defined as

$f(x) = log_{b}(x)$

where b > 0 and b ≠ 1 and also x >0, reads as log base b of x.

⇒Generally we use the base 10 i.e $\log_{10}$

⇒if $log_{b}(a) = x$ ,in exponent form :

$\implies \: b {}^{x} = a$

Solution :

we have to prove log(m×n) =log m +log n

_________________________

Let x=$\log_{10}(m)$

and y =$\log_{10}(n)$

In exponent form

$\implies m = 10 {}^{x} \: and \: n = 10 {}^{y}$

Now mutiply m and n

$mn = 10 {}^{x} \times 10 {}^{y}$

$\implies \: mn = 10 {}^{x + y}$

Now take log on both sides

$log(mn) = log(10 {}^{x + y} )$

$\implies \: log(mn) = (x + y) log(10)$

$\implies log(mn) = x + y$

$\implies log(mn) = log(m) + log(n)$

Hence proved !

Note :

Here we considered log or $\log_{10}$ both are same .