Latest Elitmus Aptitude Question SOLUTION: log6 (x^2-6x+9)^1/2 +log6(x^2-8x+16)^1/2=1 Find the no of real solution. a.)4 b.)2 c.)3 d.)0 note:-here log of base 6

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(#M40037857) ELITMUS Numerical Ability question no of real solution Keep an EYE

log6 (x^2-6x+9)^1/2 +log6(x^2-8x+16)^1/2=1
Find the no of real solution.
a.)4 b.)2 c.)3 d.)0
note:-here log of base 6

Asked In Elitmus nageshwarnk (9 years ago)

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dheeraj  (9 years ago)

by solving,we get
log6((x-3)(x-4))=1
when x=6&x=1 it will satisfy the equation
hence it will have 2 real solution

Is this solution Helpfull? Yes (21) | No (3)  

HARISHANKAR YADAV  (9 years ago)

from above eqn: log6((x-3)^2(x-4)^2)=2
log6{(x-3)(x-4)}^2 }=2
{(x-3)(x-4)}^2 }=36
(x-3)(x-4)=+6
x^2-7x+6=0
gives x=1 nd x=6
and (x-3)(x-4)=-6
x^2-7x+18=0
discrimant

Is this solution Helpfull? Yes (4) | No (3)  

HARISHANKAR YADAV  (9 years ago)

rest part
above eqn give imgnry root..left them
nw we have x=1 and x=6
it also nt satisfied orignal eqn
log6 (x^2-6x+9)^1/2 +log6(x^2-8x+16)^1/2=1
ans=d

Is this solution Helpfull? Yes (2) | No (3)  

Pooja bala singh  (9 years ago)

By Solving this we get,
log((x-3)(x-4))=log6
so, (x-3)(x-4)=6

above is clearly showing that x will have 2 solution

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Mani kandan  (9 years ago)

in place of 9x ,it was 18x.
by solving both the eq,we will get
1/2(log6(x^2-8x+16)+1/2(log6(x^2-18x+81)=1
1/2(log6(x^2-8x+16)+(log6(x^2-18x+81)))=1 , takng 1/2 common
log6(x^2-8x+16)+log6(x^2-18x+81)=2
log6((x^2-8x+16)(x^2-18x+81))=2 , acc to log rule log a + log b = log(a+b)
log6((x-4)^2(x-9)^2)=2
log6((x-4)(x-9))^2=2
2log6((x-4)(x-9))=2
log6((x-4)(x-9))=1
>>log6((x-4)(x-9))=1.....(i)
>>log6((x-4)(x-9))=log6(6)
>>(x-4)(x-9)=6
by solving..x=3,10 ans-b
it will have only 2 solution

Is this solution Helpfull? Yes (1) | No (3)  

kundan prasad  (8 years ago)

Ans:b)2
Sol: log6(x-3)(x-4)=1
after solve ,we get
36x^2-252x-422=0
D=252^2-4*36*422>0
so D>0
hence it has two 2 real solution

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