# Ln x/y = ln x - ln y From Integral 1/x dx only ?

#### morrobay

Gold Member
Is the derivation of this relationship ln x/y = ln x - ln y
exclusively from, and originating from the evaluation of the Integral from 1 --> x 1/t dt = ln x ?
To say another way can ln x/y = ln x - ln y always be considered to apply to this integral ?

#### Mentallic

Homework Helper
Not really, there are other ways to show it's true, such as using basic exponential formulae:

$$e^{x-y}=\frac{e^x}{e^y}$$

#### AlephZero

Homework Helper
Yes you can show it direct from the integral.
$$\int_1^x\frac{dt}t - \int_1^y\frac{dt}t = \int_y^x\frac{dt}t$$
Then substitute u = t/y.

#### morrobay

Gold Member
Yes you can show it direct from the integral.
$$\int_1^x\frac{dt}t - \int_1^y\frac{dt}t = \int_y^x\frac{dt}t$$
Then substitute u = t/y.
Thanks to both of you . Im not questioning the validity of the correspondence to
the integral: 1 to x 1/x dx = ln x - ln 1
Rather confirmation that this correspondence is one to one .
That ln x is derived from the above integral alone.

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#### Char. Limit

Gold Member
I'm really not sure what you mean. Do you mean that ln(x) is the only function f(x) that satisfies the property f(x/y) = f(x) - f(y)? Or do you mean that the integral is the only way of proving this property? Or perhaps something else entirely?

#### Office_Shredder

Staff Emeritus
Gold Member
I think he means don't use the fact that you know ln(x) is an inverse of ex

#### morrobay

Gold Member
Im trying to show that this property : ln x/y = ln x - ln y
is derived from the integral 1 to x 1/x dx

see this http://www.eoht.info/page/S+=+k+ln+W
I want to show that the natural logarithm in the statistical mechanics formulation
of entropy change , delta S = k ln W2/W1
is based on the natural logarithm for thermodynamic isothermal gas expansion:
delta S = nR integral V1 to V2 dv/V = ln V2/V1
That is based on the integral 1 to x 1/x dx = ln x - ln 1

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#### 6.28318531

You can prove ln(x/y) = ln(x)-ln(y), its similar to proving ln(xy)=ln(x)+ln(y), but you would use a different substitution. Start with the integral between 1 and x/y, and then split it into two integrals one between x and 1 AND the other between x/y and x. Then for the second integral use the substitution t=u/x,