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What power would I need to raise $4^{2^{2^l}}$ to get $4^{2^{2^n}}$ where $n>l$?
Very simple question but it has stumped me.

I guess $2^{2^{n-l}}$ but I do not think that is right I am not sure?

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Let p be the power.. $$(4^{2^{2^l}})^p = (4^{2^{2^n}})$$ $$p(2^{2^l})=2^{2^n}$$ $$p=2^{2^n-2^l}$$

It stops there.. you can't substract the power.. just plugin some numbers e.g. n = 3, l=1

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When you're raising to an exponent you multiply the exponents so you need to solve:

$(2^{2^l})\times p=(2^{2^n})$

taking log base 2:

$2^l + log_2 {p}=2^n$

$p=2^{2^n-2^l}$

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