HOT Competition 31 Aug/8AM: What is the sum of all possible values of

Question

What is the sum of all possible values of x for which  (-3)^{(x^2-2x-3)}=27^{(x+7)} ?

A. -3
B. 0
C. 3
D. 5
E. 8

kush09 · Answer

Solution:
=> (−3)^(x^2−2x−3)=27^(x+7)
=> (−3)^(x^2−2x−3)=3^3(x+7)
=> -3^even, then only sign will; match on both side.

=> x^2−2x−3 = 3(x+7)
=> x^2−5x−24=0
=> x = -3,8

=> 8 cannot be the solution as with 8, (-3) power will be odd.

Answer is x = -3. Option A.

Kinshook · Answer

What is the sum of all possible values of x for which  (-3)^{(x^2-2x-3)}=27^{x+7}  ?

(-3)^{(x-3)(x+1)}=3^{3(x+7)} 
If x is odd,(x-3)(x+1) is even
x^2-2x-3=3x+21
x^2-5x-24=0
(x-8)(x+3)=0
x=-3 is a valid solution.
But if x is even, (x-3)(x+1) is odd
But no solution is possible since -3^{x^2-2x-3} = 3^{3x+21}
3^{x^2-2x-3} + 3^{3x+21} =0
Both terms are always positive and sum is never 0.

x=-3 is the only valid solution
Sum of all possible values of x=-3

IMO A

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neetib · Answer

(−3)^(x2−2x−3)=27^(x+7)
we can write 27 as 3^3
equating both the powers,
x^2-2x-3=3x+21 
x^2-8x+3x-24=0
x^2-5x-24=0
x^2-8x+3x-24=0
we get x=8 and x=-3
Now right side is a positive number and the left side is a negative 
so it means (-3) should be raised to power an even number 
we will first put x=8 in x^2-2x-3 result is odd .Therefore reject 8
Now put x=-3 in x^2-2x-3 result is even . 
So OA is -3

Varunsawhney8 · Answer

What is the sum of all possible values of x for which ?

We need to make sure that we need to make even power in the LHS so as to remove the negative sign from the equation

Equating both sides
We get
x^2-2x-3=3(x+7)
x^2-2x-3-3x-21=0
x^2-5x-24=0
x=8,-3
Now putting the value in LHS to make sure power is even which will satisfy the equation

x^2-2x-3

Putting x=8 
64-16-3=45
This will not satisfy the equation

Putting x=-3
9+6-3=12

This will satisfy the equation
Hence only one value is satisfying the equation
I.e. -3
Sum is -3

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YuvarajUmapathi · Answer

What is the sum of all possible values of x for which (−3)^(x^2−2x−3)=27^(x+7)?

x^2−2x−3 must be even to hold the above equation

x^2−2x−3 = 27^(x+7)

Solve for X, we get x = 8, -3

If x=8, then (x^2−2x−3) becomes odd, equation will break.

so X=-3

sum of all possible values is -3

Answer : A

skg315 · Answer

(-3)^(x^2-2x-3) = 27^(x+7) can be written as (-1)^(x^2-2x-3)×(3)^(x^2-2x-3) = 3^(3x+21)

as 27=3^3

equating the powers of 3 from both sides

x^2-2x-3=3x+21

hence x= 8 ,-3

but we know only even power of (-1) can make LHS positive so value of x which makes (-1)^(x^2-2x-3) positive is x= -3 i.e. (-1)^12
whereas x=8 gives value (-1)^45 which will be negative

hence option A is correct

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Superman249 · Answer

IMO Ans is A

=> (−3)^ (x^2−2x−3) =27^ (x+7)   Sum of all possible value of x ?

=>  (-1)^(x^2−2x−3)  *(3)^(x^2−2x−3)=3^(3(x+7))

let us assume (x^2−2x−3) will result in Even value , as then only we can equate Power of 3------------------(1)

=> x^2−2x−3= 3(x+7)
=>x^2-5x-24=0

=> (x+3)(x-8)=0

We have two value of x=-3;  x=8

=> Lets Check value of (x^2-2x-3) for each value i.e. -3 & 8

x=-3

x^2-2x-3=>3^2+6-9=> 6  (Even)

x=8

x^2-2x-3=>8^2-16-3=> 45 (odd)

As Only x=-3 gives even no.; as per concept (1) ;   only x=-3 is acceptable

Hence, Sum of all acceptable value of x=-3

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HOT Competition 31 Aug/8AM: What is the sum of all possible values of

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HOT Competition 31 Aug/8AM: What is the sum of all possible values of

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