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If x 3/4(log 3 X)^2 + log 3 X - 5/4 = root 3. then x has a) one negative integral value b) two irrational value c) two positive rational value d) none of these If x3/4(log3X)^2 + log3X - 5/4 = root 3. then x has a) one negative integral value b) two irrational value c) two positive rational value d) none of these
If x3/4(log3X)^2 + log3X - 5/4 = root 3. then x has
a) one negative integral value
b) two irrational value
c) two positive rational value
d) none of these
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
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