Evaluate Integral of ln|x|/2 from 1 to 9: ln3 Calculation and Explanation

[tony873004]

$$\int_1^9 {\frac{1}{{2x}}\,dx}$$

$$F(x) = \frac{{\ln \left| x \right|}}{2}$$

$$\frac{{\ln \left| 9 \right|}}{2} - \frac{{\ln \left| 1 \right|}}{2} = \frac{{\ln \left| 9 \right|}}{2} = \ln 3$$
The answer ln3 came from the back of the book. I realize from using my calculator that ln9 / 2 = ln3 , but I'm not sure why. I guess I forgot my rules of ln.

Also, since the anti-derivate section gives 1/x as ln abs(x), was I correct in carrying the absolute value brackets to the ln abs(9) / 2 ? Why did the back of the book drop the absolute value brackets from the answer?

Thanks!

 

[Dick]

Because the absolute value of 3 is 3.

 

[Dick]

Also ln(9)=ln(3^2)=2*ln(3), since ln(a^b)=b*ln(a).

 

[radou]

Recall that a*ln(x) = ln (a^x).

Edit: ooops, late.

 

[tony873004]

That makes sense. I will only need the abs brackets around a variable. If it is a fixed number, just get rid of the minus sign if any, and the brackets...

Thanks... the toughest part about this calculus is remembering the all the pre-calc!

 

[neutrino]

Recall that a*ln(x) = ln (a^x).

Edit: ooops, late.

Actually, I recall it as a*ln(x) = ln(x^a).