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

$\frac{10^{5}}{ln 10}$ $10^5/ln10$ .

$β = 10 {log}_{10} (\frac{I}{10^{- 16}}) = 10 {log}_{10} (10^{16} I)$ $beta=10log_10 (I/10^-16)=10log_10(10^16I)$ .

Using the Usual Rules of $log$ $log$ functions,

$β = 10 {{log}_{10} 10^{16} + {log}_{10} I}$ $beta=10{log_10 10^16+log_10I}$ ,

$= 10 {16 {log}_{10} 10 + {log}_{10} I}$ $=10{16log_10 10+log_10 I}$ ,

$= 10 {16 + {log}_{10} I}$ $=10{16+log_10 I}$ .

$= 160 + 10 {log}_{10} I$ $=160+10log_10I$ .

Here, by the Change of Base Rule, ${log}_{10} I = \frac{ln I}{ln 10}$ $log_10I=lnI/ln10$

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

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

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

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

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

$=10^5/ln10$ .