please prove the equation....
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x^(-1) = 1/x
both side take log :-
log [x^(-1)] = log [1/x]
Using log properties :
log (a^b) = b log (a)
log a/b = log a - log b
log [x^(-1)] = log [1/x]
(-1) log (x) = log 1 - logx
- log x = 0 - logx
(we know the value of log 1 = 0)
- log x = - log x
so the
log x = log x
LHS = RHS
proved !!!! ❤️❤️❤️
both side take log :-
log [x^(-1)] = log [1/x]
Using log properties :
log (a^b) = b log (a)
log a/b = log a - log b
log [x^(-1)] = log [1/x]
(-1) log (x) = log 1 - logx
- log x = 0 - logx
(we know the value of log 1 = 0)
- log x = - log x
so the
log x = log x
LHS = RHS
proved !!!! ❤️❤️❤️
Answered by
1
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refer to the attachment above for ur answer
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