log5+log(x+10)-1=log(21x-20)-log(2x-1)

log5+log(5x+1)=log(x+5)+1

log(x-1)+log(x-2)lt log(x+2)

2log x-log(x+1)-log(x-1)=

If higher powers of x^(2) are neglected, then the value of log(1+x^(2))-log(1+x)-log(1-x)=

If higher powers of x^(2) are neglected, then the value of log(1+x^(2))-log(1+x)-log(1-x)=

If log_(10)5+log_(10)(5x+1)=log_(10)(x+5)+1, then x is equal to

int(log(x+1)-log x)/(x(x+1))dx= (A) log(x-1)log x+(1)/(2)(log x-1)^(2)-(1)/(2)(log x)^(2)+c (B) (1)/(2)(log(x+1))^(2)+(1)/(2)(log x)^(2)-log(x+1)log x+c (C) -(1)/(2)(log(x+1)^(2))-(1)/(2)(log x)^(2)+log x*log(x+1)+c (D) [log(1+(1)/(x))]^(2)+c

Solve for x: a) (log_(10)(x-3))/(log_(10)(x^(2)-21)) = 1/2 b) log(log x)+log(logx^(3)-2)= 0, where base of log is 10. c) log_(x)2. log_(2x)2 = log_(4x)2 d) 5^(logx)+5x^(log5)=3(a gt 0), where base of log is 3. e) If 9^(1+logx)-3^(1+logx)-210=0 , where base of log is 3.