Exercise
Simplify the expression:
\frac{2^{-n+1}\cdot 4^{n-1}+2^n}{2^{n+3}-2^{n-1}}
Final Answer
Solution
Using Powers and Roots rules we get:
\frac{2^{-n+1}\cdot 4^{n-1}+2^n}{2^{n+3}-2^{n-1}}=
=\frac{2^{-n+1}\cdot {(2^2)}^{n-1}+2^n}{2^n\cdot 2^3-2^n\cdot 2^{-1}}=
=\frac{2^{-n+1}\cdot 2^{2n-2}+2^n}{2^n\cdot 2^3-2^n\cdot 2^{-1}}=
=\frac{2^{-n+1+2n-2}+2^n}{2^n(2^3-2^{-1})}=
=\frac{2^n\cdot 2^{-1}+2^n}{2^n(2^3-2^{-1})}=
=\frac{2^n(2^{-1}+1)}{2^n(2^3-2^{-1})}=
=\frac{2^{-1}+1}{2^3-2^{-1}}=
=\frac{\frac{1}{2}+1}{8-\frac{1}{2}}=
=\frac{\frac{3}{2}}{\frac{15}{2}}=
=\frac{3}{2}\cdot\frac{2}{15}=
=\frac{6}{30}=\frac{1}{5}