1. May 24, 2013

### Justabeginner

1. The problem statement, all variables and given/known data
The problem asks me to integrate this: $∫(1/x) * (e^{-2 log [3, x]})) dx$
I don't know how to write log base 3 x, so that is how I entered it. Sorry if that was confusing. :S

2. Relevant equations

3. The attempt at a solution

The correct answer is $(-log(3))/(2x^{2/(log 3)}) + C$

But I have no clue how that was even the answer. Guidance is much appreciated. Thank you.

2. May 24, 2013

### CAF123

Rewrite $\log_3 x$ in terms of the natural logarithm: $$\log_ax = \frac{\ln x}{\ln a}$$

3. May 24, 2013

### Justabeginner

So I'd get int (1/x) * (e^(-2 lnx/ln3)) dx? And is the integral of e to any power that expression times the exponent?

4. May 25, 2013

### Staff: Mentor

Only if the exponent is a constant (i.e., does not contain any x terms).

You'll simplify things by using the fact that

eloge(x) = x

5. May 25, 2013

### CAF123

No. You have this $$\int \frac{1}{x} \exp \left(-2 \frac{\ln x}{\ln 3}\right)\,dx$$ There is an obvious substitution you can make now.

6. May 25, 2013

### Justabeginner

I'm not sure if you can take the integral of 1/x and multiply it by the integral of the ln part?

7. May 25, 2013

### SammyS

Staff Emeritus
No, you can't.

You have to use some additional properties of logarithms and exponential functions to get a somewhat simple integrand.

I say somewhat simple because it will just be x to a power -- but the power is pretty weird looking.

8. May 25, 2013

### Justabeginner

Oh! I see. Thank you.

9. May 25, 2013

### vela

Staff Emeritus
Since 1/x is the derivative of ln x, you might try the substitution u = ln x as well.