The problem is as follows:
The expression from below represents a model made by satellite based on the weather patters in Keelung
$$T=\sqrt[81^{3^{n}}]{\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}}$$
Assuming in the Central Weather Bureau the value of $4T$ in Celsius scale represented the temperature of Keelung at $\textrm{6:00 a.m}$. Find the temperature which will be given at $\textrm{2:00 p.m}$ using the information given by the CWB which reports that at that time the temperature will be increased $7°C$ with respect that of $\textrm{6:00 a.m}$.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&17°C\\ 2.&13°C\\ 3.&11°C\\ 4.&15°C\\ \end{array}$
From what I'm understanding in this problem the whole thing is get the value of $T$, so as with that you can get $4T$, but for this to happen it is required to simplify the expression given in the heading.
$$T=\sqrt[81^{3^{n}}]{\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}}$$
For this I'm getting:
$\left(\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}\right)^{3^{4^{3^{n}}}}$
To attack this part I concentrated in what's inside the bracket hence:
$8^{3^{3^{n+1}}}=8^{3^{3^n\cdot 3}}$
$\left(8^{3^{3^n\cdot 3}}\right)^{\frac{1}{3}}=8^{3^{3^n}}$
Since it is mentioned
$\left[2^{3^{3^{3^n}}}\right]^{3^{3^{n}}}=2^{3^{\left({3^{3^n}}\cdot 3^{3^{n}}\right)}}$
Here I'm assuming the number of nested radical would match both $3^{3^n}$ inside and outside the parenthesis, there are three numbers out so it will cover the same length from top to bottom, but is this rationale correct?.
This is reduced to (noting this part is only the exponent):
$$2^{3^{\left({3^{2\cdot 3^n}}\right)}}$$
But that's how far I went with the reduction. From then on I came stuck. Can someone help me here? which sort of manipulation should be done to solve this problem?. All that I know so far is they ask $4T$.
$$T=\sqrt[81^{3^{n}}]{\left[\sqrt[3]{8^{3^{3^{n+1}}}}\right]^{3^{3^{n}}}} = \left(\left(\left(8^{3^{3^{n+1}}}\right)^{3^{-1}}\right)^{3^{3^{n}}}\right)^{81^{-3^{n}}}$$
Looking at the exponent of $8$, we have:
$$(\log_8 T = ) \quad 3^{3^{n+1}}\cdot 3^{-1}\cdot 3^{3^n} \cdot 81^{-3^n} = 3^{3^{n+1} -1+3^n-4\cdot 3^n} = 3^{-1}$$
Hence $T = 8^{3^{-1}} = 2$.