" 32."log(log i)=

I=int e^(x log x)(1+log x)

(i) Show that (log 9)/2 + 2 log 6 +(log 81)/4 - log 12 = 3 log 3 .

int 32x^(3).(log x)^(2) dx =

State, true or false : (i) log 1 xx log 1000 = 0 (ii) (log x)/(log y) = log x - log y (iii) If (log 25)/(log 5) = log x , then x = 2 (iv) log x xx log y = log x + log y

"F i n d"int[log(logx)+1/((logx)^2)]dx Find log (log x)+ |dx 2 (log x)

Solve for x : (i) (log 81)/(log 27) = x (ii) (log 128)/(log 32) = x (iii) (log 64)/(log 8) = log x (iv) (log 225)/(log 15) = log x

(sin (log i^(i)))^3 + (cos (log i^(i)))^3 =

The real part of log log i is

Assuming the base as 10,prove that 20 log 8+45 log 16-18log 32=30(5-5log 5)

(1/2)^(logx^2)+2> 3.2^(-"log"(-x))