I am at a loss on how to start this derative problem? for calc

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Answer 1 · 2018-07-10T18:26:11

# 10^5/ln10#.

#beta=10log_10 (I/10^-16)=10log_10(10^16I)#.

Using the Usual Rules of #log# functions,

#beta=10{log_10 10^16+log_10I}#,

#=10{16log_10 10+log_10 I}#,

#=10{16+log_10 I}#.

#=160+10log_10I#.

Here, by the Change of Base Rule, #log_10I=lnI/ln10#

# rArr beta=160+10/ln10*lnI....................(ast)#.

Now, the reqd. rate#=[(dbeta)/(dI)]_(I=10^-4)#

From #(ast), (dbeta)/(dI)=0+10/ln10*1/I#.

#:."The Reqd. Rate"=[10/ln10*1/I]_(I=10^-4)#,

#=10/ln10*1/10^-4#,

#=10^5/ln10#.