<P> This can be demonstrated by splitting the integral that defines ln (ab) into two parts and then making the variable substitution x = ta in the second part, as follows: </P> <Dl> <Dd> ln ⁡ (a b) = ∫ 1 a b 1 x d x = ∫ 1 a 1 x d x + ∫ a a b 1 x d x = ∫ 1 a 1 x d x + ∫ 1 b 1 a t d (a t) = ∫ 1 a 1 x d x + ∫ 1 b 1 t d t = ln ⁡ a + ln ⁡ b . (\ displaystyle (\ begin (aligned) \ ln (ab) = \ int _ (1) ^ (ab) (\ frac (1) (x)) \, dx& = \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx+ \ int _ (a) ^ (ab) (\ frac (1) (x)) \, dx \ \ (5pt) & = \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx+ \ int _ (1) ^ (b) (\ frac (1) (at)) \, d (at) \ \ (5pt) & = \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx+ \ int _ (1) ^ (b) (\ frac (1) (t)) \, dt \ \ (5pt) & = \ ln a+ \ ln b. \ end (aligned))) </Dd> </Dl> <Dd> ln ⁡ (a b) = ∫ 1 a b 1 x d x = ∫ 1 a 1 x d x + ∫ a a b 1 x d x = ∫ 1 a 1 x d x + ∫ 1 b 1 a t d (a t) = ∫ 1 a 1 x d x + ∫ 1 b 1 t d t = ln ⁡ a + ln ⁡ b . (\ displaystyle (\ begin (aligned) \ ln (ab) = \ int _ (1) ^ (ab) (\ frac (1) (x)) \, dx& = \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx+ \ int _ (a) ^ (ab) (\ frac (1) (x)) \, dx \ \ (5pt) & = \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx+ \ int _ (1) ^ (b) (\ frac (1) (at)) \, d (at) \ \ (5pt) & = \ int _ (1) ^ (a) (\ frac (1) (x)) \, dx+ \ int _ (1) ^ (b) (\ frac (1) (t)) \, dt \ \ (5pt) & = \ ln a+ \ ln b. \ end (aligned))) </Dd> <P> In elementary terms, this is simply scaling by 1 / a in the horizontal direction and by a in the vertical direction . Area does not change under this transformation, but the region between a and ab is reconfigured . Because the function a / (ax) is equal to the function 1 / x, the resulting area is precisely ln (b). </P>

Ln 1 has the same value as 1