Product rule of Logarithms

`log_a X + log_a Y = log_a (X*Y)`

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The product rule of Logarithms

Provided that `X > 0, Y > 0` and `a ne 1` and `n, m` is any real number, then
`log_a X + log_a Y = log_a (X*Y)`

Proof

let `X = a^n` and `Y = a^m`
From the definition of logarithms, it is true that:
⇒ `log_a X = n` and `log_a Y = m`
Now,
`log_a (X*Y) = log_a (a^n * a^m)` - By substituting the values of `X` and `Y`
⇒ `log_a (X*Y) = log_a a^(n + m)` - Product rule of exponents
⇒ `log_a (X*Y) = n + m`
Now substitute the values of `n` and `m`
⇒ `log_a (X*Y) = log_a X + log_a Y`
∴ Hence proved.

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The product rule of Logarithms

Provided that `X > 0, Y > 0` and `a ne 1` and `n, m` is any real number, then
`log_a X + log_a Y = log_a (X*Y)`

Proof

let `X = a^n` and `Y = a^m`
From the definition of logarithms, it is true that:
⇒ `log_a X = n` and `log_a Y = m`
Now,
`log_a (X*Y) = log_a (a^n * a^m)` - By substituting the values of `X` and `Y`
⇒ `log_a (X*Y) = log_a a^(n + m)` - Product rule of exponents
⇒ `log_a (X*Y) = n + m`
Now substitute the values of `n` and `m`
⇒ `log_a (X*Y) = log_a X + log_a Y`
∴ Hence proved.

Recommended Lessons