ArunKumar   11 years AGO

x^2-4x-log a(base3)=0
for real, b^2-4ac=0
4^2-4*( -log a(base3)= 0
16+4*log a(base3)= 0
4*log a(base3)= -16
log a^4(base3)= -16
a^4= 3^-16
a^4= (3^-4)^4
a^4= (1/3^4)^4
a = 1/3^4
a= 1/81

Like? Yes (16) | No (4)  

ABHISHEK KUMAR MISHRA   11 years AGO

For real root (-4)^2 - -4*1*(-log a(base3))>=0
or, 16+log a(base3)>=0
16>=-log a(base3)...(1)
so here least value of a according to the option 1/81 which satisfied (1)
so option is 9)

Like? Yes (9) | No (4)  

Grishma Shah   11 years AGO

If roots of the equation are real then b^2 -4ac >=0
Here a=1 , b=-4 ,c =-loga(base 3)
b^2-4ac =16 + 4 loga
for real roots 16 + 4log a>=0
thus loga >= -4
for base 3, to satify above condition a=1/81 ,1/9,1/3
thus for minimum value of a ,answer is 1/81

Like? Yes (5) | No   

Ashfaque Hussain   11 years AGO

D=0, roots are real
16 + 4*1*log a(base3)=0
log a(base3)= -4
a = 3^-4
a=1/3^4
a=1/81

Like? Yes (6) | No (1)  

md shoaib alam   11 years AGO

a=(1/3)^4

after computing roots of equation

Like? Yes (7) | No (2)  

Rituparna Bhattacharyya   11 years AGO

ans. 9) 1/81

x^2 -4x - log a(base3)=0
=>log a(base3) = x^2 -4x
=> a = 3^(x^2-4x)
for x = 2
we have
a = 3^(-4)
thus
a = 1/81

Like? Yes (4) | No (1)  

Rats   11 years AGO

given equation :
x^2-4x-log a(base3)=0
here a=1, b=-4, c=-log a(base3)
for roots to be real
b^2-4ac=0
(-4^2)-4*1*(-log a(base3)) = 0
16+4log a(base3)=0
log a(base3)=-4
a=3^-4
a=1/81........Ans

Like? Yes (5) | No (2)  

Mayuresh Srivastava   11 years AGO

for real roots D>=0
i.e. b^2-4ac>=0
16+4log a(base3)>=0
a>=1/3^4
so least value=1/81.

Like? Yes (4) | No (2)  

kumar sundram   11 years AGO

for real root, b^2-4*a*c greater than equal to zero.
i.e, (-4)^2-4*{1*[-log a(base3)]}=0
or, 16+4*log a(base3)=0
or, 16=-4*log a(base3)
or, -4=log a(base3)
or, a=(3)^(-4)=1/81

Like? Yes (3) | No (1)  

Md Shahnawaz Shahab   11 years AGO

ans: option 9 : 1/81

x^2 -4x - log a(base3)=0
=>log a(base3) = x^2 -4x
=> a = 3^(x^2-4x)
substituting the value of 1/81 as 'a' we get
real solutions and its the least the given option
thus,
a = 1/81

Like? Yes (3) | No (2)  

Dharmendra Kumar Rajak   11 years AGO

Given equation is x^2-4x-log a(base3)=0,
is real
So for real root
a>0 and D>=0;
D=b^2-4ac
=(-4)^2-(4*1*-log a)
i.e. => 16+4 log a(base3)>=0
=> 4 log a(base3)>=-16
=> log a(base3)>=-4
=> a>=3^-4
=> a>=1/3^4
=> a>=1/81
So least value of a =1/81
So option (9) is correct

Like? Yes (1) | No   

Sriparna Boote   11 years AGO

As it is given the roots are real so D(discriminate) must be >=0.
D=b^2-4ac>=0
16-4loga(base 3)>=0
Solving we get a>=1/81
so Minimum value of a is 1/81

Like? Yes (1) | No   

ak   11 years AGO

x^2-4x-loga(base3)=0
roots are,x=-b(+-)root(b^2-4ac)/2a
x=2(+-)root(4+log3a)
if roots are real,
4-log3a>=0
a>=3^-4
a>=1/81
a=1/81

Like? Yes (1) | No   

raghuram dharmavarpu   11 years AGO

the value 81 will be the least value for above expression a=1/81
because it can be written as log a(base3)=-4
sub in above expression we get (x-2)^2 expression

Like? Yes (1) | No   

Rahul Gupta   11 years AGO

x^2-4x-log(base3)=0
x=4+-Sqrroot(16+4loga(base3))
x=2+-sqrroot(4+loga(base3))
now calculate loga(base3) you find 1/81 is less value for a.

Like? Yes (2) | No (1)  

akash   11 years AGO

For least value of a,discriminant=0
4^2+4loga(base 3)=0
solving this,value of a=3^(-4)=1/81

Like? Yes (1) | No   

jagdish sharma   11 years AGO

when roots are real then
b*b-4ac>=0
4^2-4*log(base3)a>=0
a>=1/81

Like? Yes (1) | No   

lakshay bansal   11 years AGO

x^2-4x-loga(base3)=0
b^2-4ac=0
16+4loga(base 3)=0
log a(base 3)=-4
a=3^(-4)
a=1/81

Like? Yes (1) | No   

Anantharaman   11 years AGO

Minimum value for a,

Using elimination method

Starting from minimum value for a

a = 1/81 => x^2-4x+4 = 0 => x= 2

1/81 (Option 9)

Like? Yes (2) | No (1)  

Prakash kumar   11 years AGO

For roots to be real
D = b^2 -4ac =0
16-4*1*-loga(base3)=0
16+4/loga(base3)= 0
loga(base3) = -4
a = 3^-4
a = 1/81

Like? Yes (1) | No   

Jyothi   11 years AGO

(b^2)-4ac=0
c=4 so loga(base3)=4
a=1/81

Like? Yes (1) | No   

prakash kumar   11 years AGO

for real roots
16+4loga(base3)>=0
loga(base3)>=-4
3^-4=a
a=1/81

Like? Yes (2) | No (1)  

Bharat Bhushan Jha   11 years AGO

a=1/27 and 1/81
both satisfy the given equation
but least value of a=1/81

Like? Yes (2) | No (1)  

Shrikant kumar   11 years AGO

D=B^2-4ac=0

Like? Yes (1) | No   

Saurabh Mishra   11 years AGO

4*4-4*1*(-loga(base 3))>=0
on solving this equation we get value
of a>=1/81

Like? Yes (1) | No   

sandipan patra   11 years AGO

as roots are real,
4^2-4*1*(-loga(base3))>=0
=> log a(base 3)>=-4
=> a>=1/81
least value of a=1/81

Like? Yes (1) | No (1)  

HARI om kashyap   11 years AGO

we need to find b^2-4ac>=0
4ac= 4*1*-log a base3
applying 1/81 since 81 is 4th power of 3
-log 1/81 base 3= -log 1 base 3 + log 81 base 3
=0+4
=4
d= 4^2 - 4*1*4
d=0
hence real roots

Like? Yes (1) | No (1)  

rohit kumar   11 years AGO

given,
x^2 - 4x - loga(base 3) = 0
the discriminant will be,
d = (-4)^2 - 4(1)(-log(a))
d = 16 + 4*log(a)
For the roots to be real, the discriminant must be non-negative(i.e >0),
16 + 4*log(a) >= 0
4*log(a) >= -16
log(a) >= -16/4
log(a) >= -4
a >= 3^(-4)
a >= 1/81
The minimum value of a is 1/81

Like? Yes (3) | No (3)  

ravi singh   11 years AGO

answer is 1/81
for least real value b^2-4ac=0

by putting the value og a,b,c we get
4^2+4*loga(base3)=0
therefore,1/a=3^4
hence a=1/81

Like? Yes (1) | No (1)  

Rashmi bharti   11 years AGO

b^2-4ac greater than or equal to 0
16+4log a(base 3)greater than equal to 0
a=1/81

Like? Yes (1) | No (1)  

Manish Kumar   11 years AGO

For x^2-4x-log a(base3)=0 to be real,
b^2-4ac >= 0, i.e, 16-4c >= 0
so, c

Like? Yes (2) | No (4)