pratham ashish
Last Activity: 15 Years ago
Hi,
just take the log base 3 of both side and put log(x)to the base 3 = some t
now the eqn will
=> {3/4(t)^2 +t - 5/4}t =1/2
by putng 0 & 1 u'll cm to now that one root is one
then rearrange the eqn like 3(t)^3 +4(t)^2 -5t -2 =0
now 1 is a root then divide it by (t-1) u'll get:
3(t)^2+7t+2=0
=> (3t+1)(t+2)=0
=> t = -2 , -1/3
=> x = (3)^1, (3)^-2 ,(3)^-1/3
=> x= 3, 1/9, 1/cube root of(3)
3 & 1/9 r rational & +ve but -2 is _ve integral value
Hence option a) & c) r correct
hav gd lk