Jtdylktuy

For any $log_a b$, $a>0$, $b>0$, and $a ne 1$. How do I work my way through multiple layers of this?

I'll assume you're talking about the "maximal" subset of the reals that makes the expression defined.

$log_4$ only accepts values in $(0,+infty)$ so $log_frac{1}{2}$ must have its image as this set. since $log_frac{1}{2} cdot$ is identical to $-log_2 cdot$, we have that the parameters are between $2^{-0}$ and $2^{-infty} = 0$ (abuse of notation), with both $1$ and $0$ not included. Thus $log_2 x$ is between $1$ and $0$, so $x in (1, 2)$.

You need $log_{1/2}{u} > 0$ and $log_{1/2}{u} = log_2(u)/log_{2}{1/2} = -log_2{u}$. So you really need $-log_{2}{u} > 0$. Because minus signs reverse inequalities this is asking for $log_{2}{u} < 0$. Taking $f(u) = 2^u$ on both sides gives $u < 1$.

Now $u = log_2{x}$ and the log must be non-negative for it to be a valid argument of $log_{1/2}{u}$, so $0 < log_2{x} < 1$. Applying $f(u) = 2^u$ on both sides gives $1 < x < 2$. That is the range of $x$ in which the iterated log is defined.

Just exhale and do it.

For $log_2 x$ to exist we need $x > 0$.

For $log_{frac 12} (log_2 x)$ so exist we need $log_2 x > 0$ so $x > 2^0 = 1$.

For $log_4(log_{frac 12} (log_2 x))$ to exist we need $log_{frac 12} (log_2 x)> 0$ which (as $0< frac 12 < 1$) we need $log_2 x < frac 12^0=1$ so $x < 2^1=2$.

So we need $1< x < 2$.

Alternatively we can so $log_a^k M = frac {log_a M}{log_a a^k} = frac{log_a M}{k}$

$log_4(log_{frac 12} (log_2 x)) = frac {log_2(log_{frac 12}(log_2 x))}{2}=frac {log_2(-log_2(log_2 x))}2$

So $-log_2(log_2 x) < 0$ so $0< log_2 x < 1$ and so $1 < x < 2$