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differentiat log(logx)
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1/(x log x)
Use chain rule.
oh i got it thanks
how do you do this question log x3
what is log x3?
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\[\log _{x} 3\]
when you differentiate it?
that is same as (Log 3 / log x) so it will be log3/(x(log x)^2)
why is it the same as (Log 3 / log x)
change of base\[\log_ax\to\log_bx\]is done by\[\frac{\log_bx}{\log_ab}\]
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for your formula a=x and b=whatever you want it to I guess. Natural log is a good choice.
differential of log x to base 'e' is just 1/x but differential of log x to base y is 1/(x log y). By a property of log, log x base y = (log x)/ log y here both logs on the rhs can be to any same base i have taken that base as e and then applied chain rule.
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