How Does the Logarithmic Equation Simplify with Large Frequency Values?

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

approaches this equation below for big values on f.

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

Could you please help me out?

Thanks in advance.

EDIT: I am sorry if this is posted to a wrong sub-forum.
 
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When f is large, you can ignore 1 and so there only remains [itex]-10 \log{(\frac{f}{f_B})^2}[/itex]. Now you can use the property [itex]\log a^n=n\log a[/itex] to get what you want.
 
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Shyan said:
When f is large, you can ignore 1 and so there only remains [itex]-10 \log{(\frac{f}{f_B})^2}[/itex]. Now you can use the property [itex]\log a^n=n\log a[/itex] to get what you want.
Oh gosh. Thank you for your help :)