Latest Elitmus Aptitude Question SOLUTION: 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1 How many real solutions of x exist?

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Arithmetic

Elitmus Question Papers:

(#M40034487) ELITMUS Numerical Ability question Quad Keep an EYE

1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
How many real solutions of x exist?

Asked In Elitmus Vivek Jain (10 years ago)

Unsolved Read Solution (10)

Is this Puzzle helpful?   (8)   (0) Submit Your Solution Arithmetic

Read Solution (10) :

RAKESH  (10 years ago)

1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1

=> 1/2[ log base 6 of (x^2-8x+16) + log base 6 of(x^2-18x+81)]= 1

=> log base 6 of (x-4)^2 + log base 6 of (x-9)^2 = 2

=> log base 6 of [(x-4)^2*(x-9)^2] = log base 6 of 36

=> (x-4)^2 * (x-9)^2 = 36

=> (x-4)*(x-9)=6 0r (x-4)*(x-9)= -6

=> x^2-13x+30=0 0r x^2-13x+42=0

=> (x-3)(x-10)= 0 or (x-6)(x-7)= 0

=> x= 3,10,6,7

all 4 values satisfies the given eqn,

4 real solns

Is this solution Helpfull? Yes (41) | No (5)  

RAKESH  (10 years ago)

1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1

=> 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81))= 1

=> log base 6 of (x^2-8x+16)^(1/2) + log base 6 of(x^2-18x+81)^(1/2)= 1

=> log base 6 of {(x-4)^2}^(1/2) + log base 6 of {(x-9)^2}^(1/2) = 1

=> log base 6 of (x-4) + log base 6 of (x-9) = 1

=> log base 6 of {(x-4)*(x-9)} = log base 6 of 6

=> (x-4)*(x-9)= 6
=> x^2-13x+30 = 0
=> (x-3)(x-10)= 0
=> x = 3,10

both x=3 & x=10 satisfies given eqn

no. of real solutions of x = 2

Is this solution Helpfull? Yes (26) | No (13)  

Vivek Jain  (10 years ago)

only one real solution exist x=10.

=>log base 6 of {(x-4)(x-9)}=1
=>{(x-4)(x-9)}>0
Because x>9 x=3 is not the solution of this equation.

Is this solution Helpfull? Yes (6) | No (11)  

shree  (10 years ago)

1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1

=> 1/2[ log base 6 of (x^2-8x+16) + log base 6 of(x^2-18x+81)]= 1

=> log base 6 of (x-4)^2 + log base 6 of (x-9)^2 = 2

=> log base 6 of [(x-4)^2*(x-9)^2] = log base 6 of 36

=> (x-4)^2 * (x-9)^2 = 36

=> (x-4)*(x-9)=6 0r (x-4)*(x-9)= -6

=> x^2-13x+30=0 0r x^2-13x+42=0

=> (x-3)(x-10)= 0 or (x-6)(x-7)= 0
so x= 3,10,6,7

and only the 4 solution is here

Is this solution Helpfull? Yes (5) | No (4)  

ritesh kumar raushan  (10 years ago)

1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1

=> 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81))= 1

=> log base 6 of (x^2-8x+16)^(1/2) + log base 6 of(x^2-18x+81)^(1/2)= 1

=> log base 6 of {(x-4)^2}^(1/2) + log base 6 of {(x-9)^2}^(1/2) = 1

=> log base 6 of (x-4) + log base 6 of (x-9) = 1

=> log base 6 of {(x-4)*(x-9)} = log base 6 of 6

=> (x-4)*(x-9)= 6
=> x^2-13x+30 = 0
=> (x-3)(x-10)= 0
=> x = 3,10
Now
=>log base 6 of {(x-4)(x-9)}=1
=>{(x-4)(x-9)}>0
for x=10
(x-4)(x-9)=6>0
for x=3
(x-4)(x-9)=-1*-6=6>0
So,
no. of real solutions of x is 2

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

Bhanu Tyagi  (10 years ago)

1/2(log base 6 0f (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81))=1

=> 1/2 ( log base 6 of (x-4)^2 + log base 6 0f (x-9)^2 )=1

=> log base 6 0f ((x-4)^2*(x-9)^2)=2

=> log base 6 of ( (x-4)^2*(x-9)^2)= log base 6 of 36

=> (x-4)^2 * (x-9)^2 = 36

So, now possible L.H.S. will be - 1 * 36 and 4 * 9 ( bec. squares
multiplication is 36 )
So, x can be x=3 , x= 6 and x=10 . So, 3 real values possible.

Is this solution Helpfull? Yes (1) | No (5)  

srinu baratam  (10 years ago)

Ans. is 4
log base 6^2 of (x^2-8*x+16)+log base 6^2 of (x^2-18*x+81)=1
log base 36 of (x^2-8*x+16)(x^2-18*x+81)=1
(x^2-8*x+16)(x^2-18*x+81)=36
(x-4)^2*(x-9)^2=6^2
(x-4)*(x-9)=-6 or (x-4)*(x-9)=+6
here x=3,10 here x=6,7
so, totally 4

Is this solution Helpfull? Yes (1) | No   

Vivek Jain  (10 years ago)

The problem here is, for guys who are saying the solutions are 2.The options were 0,1,3,4

Is this solution Helpfull? Yes | No   

harish  (10 years ago)

x>5 and x>-8

Is this solution Helpfull? Yes | No   

sonu kumar  (7 years ago)

what is exact answer ?

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.