log(a+ib)" when "a>0,b<0" is "

Solve :2log_(x)a+log_(ax)a+3log_(b)a=0 where a>0,b=a^(2)x

tan (i log ((a + ib) / / a-ib))) =

If f(x) = x (1 - log x), " where " x gt 0 , show that (a -b) log c = b (1 - logb) - a(1 - log a), " where " 0 lt a lt c lt b

tan [ i log ((a - ib)/(a + ib )) ] is equal to : a) ab b) (2 ab)/( a ^(2) - b ^(2)) c) (a ^(2) - b ^(2))/( 2 ab) d) (2 ab)/( a ^(2) + b ^(2))

If (a^(log_b x))^2-5 a^(log_b x)+6=0, where a >0, b >0 & a b!=1, then the value of x can be equal to (a) 2^(log_b a) (b) 3^(log_a b) (c) b^(log_a2) (d) a^(log_b3)

Prove that a^x-b^y=0 w h e r e x=" "sqrt(("log")_a b) & y=sqrt((log)_b a ), a >0, b >0 & a , b!=1

If ((log)_a N)/((log)_c N)=((log)_a N-(log)_b N)/((log)_b N-(log)_c N), where N >0 and N!=1, a , b , c >0,!=1 , then prove that b^2=a c

Prove that tan(ilog_e((a-ib)/(a+ib)))=(2ab)/(a^2-b^2) (where a, b in R^+ )

Prove that tan(ilog_e((a-ib)/(a+ib)))=(2ab)/(a^2-b^2) (where a, b in R^+ )