化简 1/2 ln(x)+3[ln(x-1)-2/9 ln(x+1)]

解答

化简

\frac{1}{2} ln (x) + 3 [ln (x - 1) - \frac{2}{9} ln (x + 1)]

\frac{1}{2} ln (x) + 3 ln (\frac{( x - 1 ) ( x + 1 ) ^{\frac{7}{9}}}{x + 1})

求解步骤

\frac{1}{2} ln (x) + 3 [ln (x - 1) - \frac{2}{9} ln (x + 1)]

使用对数计算法则:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{2}{9} ln (x + 1) = ln ((x + 1)^{\frac{2}{9}})

= \frac{1}{2} ln (x) + 3 [ln (x - 1) - ln ((x + 1)^{\frac{2}{9}})]

使用对数计算法则:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (x - 1) - ln ((x + 1)^{\frac{2}{9}}) = ln (\frac{x - 1}{( x + 1 ) ^{\frac{2}{9}}})

= \frac{1}{2} ln (x) + 3 [ln (\frac{x - 1}{( x + 1 ) ^{\frac{2}{9}}})]

化简

\frac{x - 1}{( x + 1 ) ^{\frac{2}{9}}} : \frac{( x - 1 ) ( x + 1 ) ^{\frac{7}{9}}}{x + 1}

= \frac{1}{2} ln (x) + 3 [ln (\frac{( x - 1 ) ( x + 1 ) ^{\frac{7}{9}}}{x + 1})]

去除括号:

(a) = a

= \frac{1}{2} ln (x) + 3 ln (\frac{( x - 1 ) ( x + 1 ) ^{\frac{7}{9}}}{x + 1})

s im pl i f y lo g_{b} (28) - lo g_{b} (7)

e x p an d lo g_{10} (\frac{z ^{5}}{y x ^{3}})

s im pl i f y 2 lo g_{10} (3 x)

s im pl i f y lo g_{10} (\frac{243}{32}) + lo g_{10} (\frac{32}{243})

s im pl i f y lo g_{10} (\frac{1}{10 x})