$T(n) = T(n^{\frac{1}{a}}) + 1$
(Change of variables method)
Put $n = 2^m$
$m = log_2n$
$T(2^m) = T(2^{\frac{m}{a}}) + 1$
Let $S(m) = T(2^m)$, which gives $S(\frac{m}{a}) = T(2^{\frac{m}{a}})$
$S(m) = S(\frac{m}{a}) + 1$
since $T(b)=1$ and we assumed $S(m) = T(2^m)$ so if $2^m$ = $b$
$T(2^m) = 1$ so $m=log_2b$
$S(log_2b) = 1$
$S(m) = S(\frac{m}{a^2}) + 1+1$
$S(m) = S(\frac{m}{a^2}) + 2$
$S(m) = S(\frac{m}{a^3}) +1 + 2$
$S(m) = S(\frac{m}{a^3}) + 3$
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$S(m) = S(\frac{m}{a^k}) + k$
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as $S(log_2b) = 1 ,\frac{m}{a^k} = log_2b$ to make$ S(\frac{m}{a^k})$ equal to 1
$m = log_2n$
$\frac{log_2n}{a^k} = log_2b$
$\frac{log_2n}{log_2b} =a^k$
make base b by base change rule
$\frac{log_bn}{log_bb} =a^k$
${log_bn} =a^k$
apply $log_a$ on both sides
$k=log_alog_bn$
$S(m) = S(\frac{m}{a^k}) + k$
$S(m) = 1+ log_a log_bn$
$ = Θ(log_a log_bn) $