solvefor x,4^{x+1}+2^y=20

解答

求解

x, 4^{x + 1} + 2^{y} = 20

x = \frac{ln ( 20 - 2 ^{y} )}{2 ln ( 2 )} - 1

求解步骤

4^{x + 1} + 2^{y} = 20

将

2^{y}

到右边

4^{x + 1} = 20 - 2^{y}

使用指数运算法则

2 (x + 1) ln (2) = ln (20 - 2^{y})

解

2 (x + 1) ln (2) = ln (20 - 2^{y}) : x = \frac{ln ( 20 - 2 ^{y} )}{2 ln ( 2 )} - 1

x = \frac{ln ( 20 - 2 ^{y} )}{2 ln ( 2 )} - 1

so l v e f or t, (1.045)^{t} = 2

so l v e f or x, 1 0^{- 6 x} = 10 0^{3 - 2 x}

3 \cdot 2^{x + 1} - 64 = 2^{x + 1}

x + 2 8 = 2 x - 1 32

(5)^{x} = 125