# Properties of Natural Logarithm $\ln(x^a)=a\ln x$

prove by using $$\ln x = \int_1^{x} dt/t$$ and without using Derivative:

$$\ln(x^a)=a\ln x \ \ \ \ x\in \mathbb{R}^+ , a \in \mathbb{R}$$

My Try :

$$\ln (x^a) = \int_1^{x^a} dt/t =$$

Now how ?

• What is $y$ doing there? Commented Jun 4, 2017 at 7:15
• @S.vanNigtevecht sorry I edited. Commented Jun 4, 2017 at 7:17
• Also, the title doesn't really match the question. $a$ and $x$ are switched...
– RGS
Commented Jun 4, 2017 at 7:17

After writing that integral, lets try the substitution

$$u^a = t$$

Then we need to change the bounds of the integral. Can you verify that the new bounds are $1$ and $x$?

After that, we also need to change the expression inside the integral. First, we find $du$:

$$u^a = t \implies au^{a-1}du = dt$$ and we can substitute everything in the integral:

$$\int_1^{x^a} \frac{dt}{t} = \int_1^x \frac{au^{a-1}du}{u^a} = a\int_1^x \frac{du}{u} = a\ln{x}$$

Let $u^a=t$. Then $au^{a-1}du=dt$.

$$\int_1^{x^a}\frac{dt}{t}=\int_1^{x}\frac{au^{a-1}du}{u^a}=a\int_1^x\frac{du}{u}=a\ln(x)$$

Doing the substitution $t=uâ$ and $dt=au^{a-1}\,du$, you get$$\int_1^{x^a}\frac{dt}t=\int_1^x\frac{au^{a-1}}{u^a}\,du=a\int_1^x\frac{du}u=a\ln x.$$