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受欢迎的代数 >

展开 log_{9}(\sqrt[7]{(s^6t)/(81)})

解答

展开

lo g_{9} (7 \frac{s ^{6} t}{81})

解答

- \frac{2}{7} + \frac{lo g _{9} ( t )}{7} + \frac{6 lo g _{9} ( s )}{7}

求解步骤

lo g_{9} (7 \frac{s ^{6} t}{81})

改写为

= lo g_{9} ((\frac{s ^{6} t}{81})^{\frac{1}{7}})

使用对数运算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

假定

x \geq 0

= \frac{1}{7} lo g_{9} (\frac{s ^{6} t}{81})

化简

lo g_{9} (\frac{s ^{6} t}{81}) : 6 lo g_{9} (s) + lo g_{9} (t) - 2

= \frac{1}{7} (6 lo g_{9} (s) + lo g_{9} (t) - 2)

打开括号

= \frac{1}{7} lo g_{9} (t) + \frac{1}{7} \cdot 6 lo g_{9} (s) + \frac{1}{7} (- 2)

使用加减运算法则

+ (- a) = - a

= \frac{1}{7} lo g_{9} (t) + 6 \cdot \frac{1}{7} lo g_{9} (s) - 2 \cdot \frac{1}{7}

6 \cdot \frac{1}{7} lo g_{9} (s) = \frac{6}{7} lo g_{9} (s)

2 \cdot \frac{1}{7} = \frac{2}{7}

= \frac{1}{7} lo g_{9} (t) + \frac{6}{7} lo g_{9} (s) - \frac{2}{7}

乘

\frac{1}{7} lo g_{9} (t) : \frac{lo g _{9} ( t )}{7}

= \frac{lo g _{9} ( t )}{7} + lo g_{9} (s) \frac{6}{7} - \frac{2}{7}

乘

lo g_{9} (s) \frac{6}{7} : \frac{6 lo g _{9} ( s )}{7}

= \frac{lo g _{9} ( t )}{7} + \frac{6 lo g _{9} ( s )}{7} - \frac{2}{7}

合并分式

\frac{lo g _{9} ( t )}{7} + \frac{6 lo g _{9} ( s )}{7} : \frac{lo g _{9} ( t ) + 6 lo g _{9} ( s )}{7}

= - \frac{2}{7} + \frac{6 lo g _{9} ( s ) + lo g _{9} ( t )}{7}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 2 + lo g _{9} ( t ) + 6 lo g _{9} ( s )}{7}

- 2 + lo g_{9} (t) + 6 lo g_{9} (s) = - 2 + lo g_{9} (t s^{6})

= \frac{- 2 + lo g _{9} ( s ^{6} t )}{7}

lo g_{9} (t s^{6}) = lo g_{9} (t) + 6 lo g_{9} (s)

= \frac{- 2 + lo g _{9} ( t ) + 6 lo g _{9} ( s )}{7}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 2 + lo g _{9} ( t ) + 6 lo g _{9} ( s )}{7} = - \frac{2}{7} + \frac{lo g _{9} ( t )}{7} + \frac{6 lo g _{9} ( s )}{7}

= - \frac{2}{7} + \frac{lo g _{9} ( t )}{7} + \frac{6 lo g _{9} ( s )}{7}

流行的例子

展开 2/(sqrt(9+4x))

e x p an d \frac{2}{9 + 4 x}

化简 (x^3-2)(sec^2(x))

s im pl i f y (x^{3} - 2) (sec^{2} (x))

化简 e^{in(sec(θ)+tan(θ))}

s im pl i f y e^{in (s e c (θ) + t a n (θ))}

展开 1/((1-e^x))

e x p an d \frac{1}{( 1 - e ^{x} )}

展开-((x+αt)^2)/2

e x p an d - \frac{( x + α t ) ^{2}}{2}