How Does the Logarithmic Equation Simplify with Large Frequency Values?

[Rectifier]

Hey!
I have a question regarding a statement in my physics book. I don't see how
|H(f)|_{dB} = - 10log (1+( \frac{f}{f_B})^2)

approaches this equation below for big values on f.

|H(f)|_{dB} = - 20log ( \frac{f}{f_B})

Could you please help me out?

Thanks in advance.

EDIT: I am sorry if this is posted to a wrong sub-forum.

 

[ShayanJ]

When f is large, you can ignore 1 and so there only remains -10 \log{(\frac{f}{f_B})^2}. Now you can use the property \log a^n=n\log a to get what you want.

 

[Rectifier]

When f is large, you can ignore 1 and so there only remains -10 \log{(\frac{f}{f_B})^2}. Now you can use the property \log a^n=n\log a to get what you want.

Oh gosh. Thank you for your help :)