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受欢迎的代数 >

64ln(2b)ln(8b)=22

解答

64 ln (2 b) ln (8 b) = 22

解答

b = e^{\frac{- 128 l n ( 2 ) + 4096 l n ^{2} ( 2 ) + 1408}{64}}, b = e^{\frac{- 128 l n ( 2 ) - 4096 l n ^{2} ( 2 ) + 1408}{64}}

+1

十进制

b = 0.61975 \dots, b = 0.10084 \dots

求解步骤

64 ln (2 b) ln (8 b) = 22

两边除以

2

\frac{64 ln ( 2 b ) ln ( 8 b )}{2} = \frac{22}{2}

化简

32 ln (2 b) ln (8 b) = 11

使用对数运算法则

32 (ln (2) + ln (b)) (ln (8) + ln (b)) = 11

用

ln (b) = u

改写方程式

32 (ln (2) + u) (ln (8) + u) = 11

解

32 (ln (2) + u) (ln (8) + u) = 11 : u = \frac{- 128 ln ( 2 ) + 4096 ln ^{2} ( 2 ) + 1408}{64}, u = \frac{- 128 ln ( 2 ) - 4096 ln ^{2} ( 2 ) + 1408}{64}

u = \frac{- 128 ln ( 2 ) + 4096 ln ^{2} ( 2 ) + 1408}{64}, u = \frac{- 128 ln ( 2 ) - 4096 ln ^{2} ( 2 ) + 1408}{64}

代回

u = ln (b)

，求解

b

解

ln (b) = \frac{- 128 ln ( 2 ) + 4096 ln ^{2} ( 2 ) + 1408}{64} : b = e^{\frac{- 128 l n ( 2 ) + 4096 l n ^{2} ( 2 ) + 1408}{64}}

解

ln (b) = \frac{- 128 ln ( 2 ) - 4096 ln ^{2} ( 2 ) + 1408}{64} : b = e^{\frac{- 128 l n ( 2 ) - 4096 l n ^{2} ( 2 ) + 1408}{64}}

b = e^{\frac{- 128 l n ( 2 ) + 4096 l n ^{2} ( 2 ) + 1408}{64}}, b = e^{\frac{- 128 l n ( 2 ) - 4096 l n ^{2} ( 2 ) + 1408}{64}}

验证解:

b = e^{\frac{- 128 l n ( 2 ) + 4096 l n ^{2} ( 2 ) + 1408}{64}}

真

, b = e^{\frac{- 128 l n ( 2 ) - 4096 l n ^{2} ( 2 ) + 1408}{64}}

真

解为

b = e^{\frac{- 128 l n ( 2 ) + 4096 l n ^{2} ( 2 ) + 1408}{64}}, b = e^{\frac{- 128 l n ( 2 ) - 4096 l n ^{2} ( 2 ) + 1408}{64}}

作图

流行的例子

x - 3 ln (x) = ln (5)

- 7 + ln (2 x) = 4

log_{10}(sqrt(x^320000))=2

lo g_{10} (x^{3} 20000) = 2

log_{7}(x)=9log_{x}(7)

lo g_{7} (x) = 9 lo g_{x} (7)

log_{10}(x^2+2x-8)=0

lo g_{10} (x^{2} + 2 x - 8) = 0