algorithm - Is (log n)^2 same as log(^2) n? -

is log(^2) n = log (log n) or (log n)^2?

i reading masters theorem , encountered video here. here, mr ravula said log(^2) n not (log n * log n), later in video, precisely @ 11:50th min, wrote n/(log n) n*(log^(-1) n). confused because: n/(log n)=n*(log n)^-1, shouldn't n*(log(^-1) n) (as per video). can please me understand difference between 2 log representations (my basics seems hazy on topic). if have equation masters theorem (t(n)=at(n/b)+f(n)) f(n) of form (log n)*(log n), can write log(^2) n or log(^2) n represent (as mr ravula told) log (log n)?

yes, mr. ravindrababu ravula right in claiming same.

here, mr ravula said log(^2) n not (log n * log n)!

he meant log(^2) n (log (log n)) causing confusion. mentioned ami tavory in comments section, log(^2) n non-standard representation. perhaps, shouldn't have been used express such expression. should avoid using representation.

but, confusing other formuals.

so confused because: n/(log n)=n*(log n)^-1, shouldn't n*(log(^-1) n) (as per video).

yes, negaitve representation of log doesn't exist in opposed positive squared powers. mentioned, cause confusion non-standard represenation.

so, can write n/(log n)=n*(log n)^-1 not n*(log(^-1) n).

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the basic formula powers of logarithms says :-

(log n)^k = log^k n.

like, x = log n, x^2 = (log n)^2 = log n * log n = log²n.

note not same log_kn brings base k.

check these links verify powers of logarithms mentioned in second half of answer ---> difference between log n , log^2 n , what big-o complexity of function (log n)^k k?.