(3^{2x+1})*(2^{x+1})=6

解答

(3^{2 x + 1}) \cdot (2^{x + 1}) = 6

x = \frac{ln ( 6 )}{ln ( 18 )} + \frac{- 1 - lo g _{2} ( 3 )}{2 lo g _{2} ( 3 ) + 1}

十进制

x = 0

求解步骤

(3^{2 x + 1}) (2^{x + 1}) = 6

使用指数运算法则

(lo g_{2} (3) (2 x + 1) + x + 1) ln (2) = ln (6)

解

(lo g_{2} (3) (2 x + 1) + x + 1) ln (2) = ln (6) : x = \frac{ln ( 6 )}{ln ( 18 )} + \frac{- 1 - lo g _{2} ( 3 )}{2 lo g _{2} ( 3 ) + 1}

x = \frac{ln ( 6 )}{ln ( 18 )} + \frac{- 1 - lo g _{2} ( 3 )}{2 lo g _{2} ( 3 ) + 1}

so l v e f or m, 144 = \frac{( 2 ^{m} )}{( 3 ^{n} )}

\frac{x}{( \frac{1}{2} )} = 2^{x} + 3

(\frac{2}{5})^{3 x - 7} = (\frac{5}{2})^{7 x - 3}

e^{x} + e^{- x} = 2

2^{n - 1} + 2^{n + 1} = 320