Question

Simplify

Answer 1

\[2^{21} / 4^{4}(4^{4}+4^{5})\]

Answer 2

@fda
we have
\[\frac{2^{21}}{4^4(4^4+4^5)}\]
we know that
\[4=2^2 \]
so
\[\frac{2^{21}}{(2^2)^4((2^2)^4+(2^2)^5)}\]
now we'll use the property
\[(x^A)^B=x^{A\times B}\]
so we get now
\[\large \frac{2^{21}}{2^8(2^8+2^{10})}\]
We can factor out 2^{8} from the bracketed term in the denominator
We have now
\[\frac{2^{21}}{2^8\times 2^8(1+2^2)}\]
Can you solve it from here?