Question

Proof logn(m×n) = lognm + lognn​

Answer 1

Question :

Proof that log(m×n) =log m +log n

Logarithm function :

The Logarithm function is defined as
f(x) = log_{b}(x)f(x)=logb​(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}log10​
⇒if log_{b}(a) = xlogb​(a)=x ,in exponent form :
\implies \: b {}^{x} = a⟹bx=a

Solution :

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

_________________________

Let x=\log_{10}(m)log10​(m)
and y =\log_{10}(n)log10​(n)
In exponent form
\implies m = 10 {}^{x} \: and \: n = 10 {}^{y}⟹m=10xandn=10y
Now mutiply m and n
mn = 10 {}^{x} \times 10 {}^{y}mn=10x×10y
\implies \: mn = 10 {}^{x + y}⟹mn=10x+y
Now take log on both sides
log(mn) = log(10 {}^{x + y} )log(mn)=log(10x+y)
\implies \: log(mn) = (x + y) log(10)⟹log(mn)=(x+y)log(10)
\implies log(mn) = x + y⟹log(mn)=x+y
\implies log(mn) = log(m) + log(n)⟹log(mn)=log(m)+log(n)

Hence proved !

brainlist (:

Answer 2

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