浙江省中考数学典型
若: n n = 2 2048 n^n=2^{2048} nn=22048求 n \sqrt{n} n
解
幂运算公式1
a n = a t × 1 t × n a^n=a^{t \times \frac{1}{t}\times n} an=at×t1×n
n n = 2 2048 = 2 t × 1 t × 2048 = ( 2 t ) 2048 t n^n=2^{2048}=2^{t\times\frac{1}{t}\times 2048}=(2^t)^{\frac{2048}{t}} nn=22048=2t×t1×2048=(2t)t2048
总有一个实数 t t t 满足 n = 2 t = 2048 t n=2^t=\frac{2048}{t} n=2t=t2048
t × 2 t = 2048 = 2 11 t\times 2^t=2048=2^{11} t×2t=2048=211
幂运算公式2
a n − m × a m = a n − m + m = a n a^{n-m}\times a^{m}=a^{n-m+m}=a^n an−m×am=an−m+m=an
t × 2 t = 2 11 = 2 11 − m × 2 m t\times 2^t =2^{11}=2^{11-m} \times2^m t×2t=211=211−m×2m
总有一个实数 m m m 满足 t = 2 m = 11 − m t=2^m=11-m t=2m=11−m
移项: 2 m + m = 11 = 8 + 3 = 2 3 + 3 2^m+m=11=8+3=2^3+3 2m+m=11=8+3=23+3
∵ 2 m + m = 8 + 3 = 2 3 + 3 \because 2^m+m=8+3=2^3+3 ∵2m+m=8+3=23+3
∴ m = 3 \therefore m=3 ∴m=3
∴ t = 2 m = 2 3 = 8 \therefore t=2^m=2^3=8 ∴t=2m=23=8
∴ n = 2 t = 2 8 \therefore n=2^t=2^8 ∴n=2t=28
∴ n = 2 8 = 2 8 2 = 2 4 = 16 \therefore \sqrt n=\sqrt{2^8}=2^\frac{8}{2}=2^4=16 ∴n=28=228=24=16