# Thread: Prove a natural logarithmic property using ... ?

1. ## Prove a natural logarithmic property using ... ?

The Question:
Use the definition of the natural logarithm function to prove that :
ln (ab) = ln(a) + ln(b)
for any positive number a and b.

my problem is that I do not know how to prove it using the definition

BTW, the definition is $\displaystyle \int_{1}^{x} \dfrac{dt}{t} = ln|x|$

2. Originally Posted by Miss
The Question:
Use the definition of the natural logarithm function to prove that :
ln (ab) = ln(a) + ln(b)
for any positive number a and b.

my problem is that I do not know how to prove it using the definition

BTW, the definition is $\displaystyle \int_{1}^{x} \dfrac{dt}{t} = ln|x|$
By use of the chain rule we get that

$\displaystyle \displaystyle\frac{d}{db}\ln(a\cdot b)=\frac{d}{db}\int_1^{a\cdot b}\frac{dt}{t}=\frac{1}{a\cdot b}\cdot a=\frac{1}{b}$

Hence, by integration, we conclude that

$\displaystyle \text{(*)}\qquad \ln(a\cdot b)=\ln(b)+c$

For some value of $\displaystyle c$. To determine the value of $\displaystyle c$ we set $\displaystyle b=1$ in (*) to find that

$\displaystyle \ln(a\cdot 1)=\ln(1)+c=0+c=c$, i.e. $\displaystyle c=\ln(a)$

Plugging this information into (*) we can finally say, with confidence, that

$\displaystyle \ln(a\cdot b)=\ln(b)+\ln(a)$

must hold.

3. Thanks
but you have a small mistake
the lower limit for integral should be 1 not 0

thanks again

4. Originally Posted by Miss
Thanks
but you have a small mistake
the lower limit for integral should be 1 not 0

thanks again
Terrible mistake, as if the lower limit of every integral were 0! - I'm going to correct that mistake right away.

5. A slight variation, a bit more direct, I think, since it uses the integral only and not the derivative:

If a is positive, then $\displaystyle \frac{1}{a}$ is also positive and, by that definition, $\displaystyle ln(1/x)= \int_1^{1/a} \frac{1}{t}dt$.

Let u= at so that t= u/a, dt= (1/a)du. When t= 1, u= a, when t= 1/a, u= 1. Now we have $\displaystyle ln(1/a= \int_{a}^1 (a/u)(1/a)du$. The "a"s cancel giving
$\displaystyle ln(1/a)= \int_a^1 \frac{1}{u} du= -\int_1^a \frac{1}{u}du= -ln(a)$.

If a and b are positive then ab is also positive and, by that definition,
$\displaystyle ln(ab)= \int_1^{ab}\frac{1}{t}dt$.

Now, let u= t/a so that t= au, dt= adu, (1/t)dt= (1/au)(adu)= (1/u)du. When t= 1, u= 1/a and when t= ab, u= b. The integral becomes $\displaystyle \int_{1/a}^b \frac{1}{u}du=$$\displaystyle \int_{1/a}^1 \frac{1}{u}du+ \int_1^b \frac{1}{u}du$$\displaystyle = -\int_1^{1/a}\frac{1}{u}du+ \int_1^b \frac{1}{u}du= -ln(1/a)+ ln(b)= ln(a)+ ln(b)$.

6. Another way.
Here is the way Leonard Gillman does this problem.
Because $\displaystyle x>0$ we use $\displaystyle \int_1^x {\frac{{du}}{u}} = \ln (x)$.

Using u-substitution $\displaystyle \int_a^{ab} {\frac{{dx}}{x}} = \int_1^b {\frac{{du}}{u}} ;\;u = \frac{x}{a}$.

Now consider that $\displaystyle \int_1^{ab} {\frac{{dx}}{x}} = \int_1^a {\frac{{dx}}{x}} + \int_a^{ab} {\frac{{dx}}{x}} = \int_1^a {\frac{{dx}}{x}} + \int_1^b {\frac{{dx}}{x}}$

7. HallsOfIvy :