\(\frac{2^n}{2}+2^2.2^n=9.2^5\)
=> \(2^{n-1}+2^{2+n}=2^5.9\)
=> \(2^{n-1}+2^{n-1+3}=2^5.9\)
=> \(2^{n-1}.\left(1+2^3\right)=2^5.9\)
=> \(2^{n-1}.9=2^5.9\)
=> \(2^{n-1}=2^5\)
=> \(n-1=5\)
Vậy n = 6.
Ta có:
\(\frac{2^n}{2}+2^2.2^n=2^{n-1}+2^3.2^{n-1}=2^{n-1}.\left(1+2^3\right)\)
\(=2^{n-1}.9=9.2^5\)
Chia cả 2 vế cho 9 thì có:
\(2^{n-1}=2^5\Rightarrow2^n=2^6\Rightarrow n=6\)