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?
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?
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:
$2^l + log_2 {p}=2^n$
$p=2^{2^n-2^l}$