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MCQEasyJEE 2024Quadratic Equations in Complex Numbers

JEE Mathematics 2024 Question with Solution

The sum of all solutions of the equation 82x−16⋅8x+48=08^{2x} - 16 \cdot 8^x + 48 = 0 is:

  • A

    1+log⁡6(8)1 + \log_{6}(8)

  • B

    log⁡8(6)\log_{8}(6)

  • C

    1+log⁡8(6)1 + \log_{8}(6)

  • D

    log⁡8(4)\log_{8}(4)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: 82x−16⋅8x+48=08^{2x} - 16 \cdot 8^x + 48 = 0

Find: The sum of all solutions.

Use the substitution y=8xy = 8^x. Then the equation becomes

y2−16y+48=0y^2 - 16y + 48 = 0

Factor the quadratic:

(y−4)(y−12)=0(y - 4)(y - 12) = 0

So,

y=4ory=12y = 4 \quad \text{or} \quad y = 12

Reverting to y=8xy = 8^x,

8x=4⇒x=log⁡8(4)8^x = 4 \Rightarrow x = \log_8(4)

and

8x=12⇒x=log⁡8(12)8^x = 12 \Rightarrow x = \log_8(12)

Therefore, the sum of the solutions is

log⁡8(4)+log⁡8(12)=log⁡8(48)\log_8(4) + \log_8(12) = \log_8(48)

Now,

log⁡8(48)=log⁡8(8⋅6)=log⁡8(8)+log⁡8(6)=1+log⁡8(6)\log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6)

Therefore, the sum of all solutions is 1+log⁡8(6)1 + \log_8(6). The correct option is C.

The first provided approach contains a typo in the final logarithm expression, but the working and the second approach correctly give 1+log⁡8(6)1 + \log_8(6).

Direct Factorisation Trick

Given: 82x−16⋅8x+48=08^{2x} - 16 \cdot 8^x + 48 = 0

Find: The sum of all solutions.

Recognize the equation as a quadratic in 8x8^x. Let t=8xt = 8^x. Then

t2−16t+48=0t^2 - 16t + 48 = 0

This factors immediately as

(t−4)(t−12)=0(t - 4)(t - 12) = 0

Hence the two values are t=4t = 4 and t=12t = 12, so the corresponding solutions are x=log⁡8(4)x = \log_8(4) and x=log⁡8(12)x = \log_8(12). Their sum is

log⁡8(4)+log⁡8(12)=log⁡8(48)=log⁡8(8⋅6)=1+log⁡8(6)\log_8(4) + \log_8(12) = \log_8(48) = \log_8(8 \cdot 6) = 1 + \log_8(6)

Therefore, the correct option is C.

Common mistakes

  • Taking 82x8^{2x} as 82⋅x8^2 \cdot x is incorrect because exponent rules give 82x=(8x)28^{2x} = (8^x)^2. Treat the equation as a quadratic in 8x8^x instead.

  • After finding 8x=48^x = 4 and 8x=128^x = 12, stopping there is incomplete because the variable asked is xx. Convert each to logarithmic form: x=log⁡8(4)x = \log_8(4) and x=log⁡8(12)x = \log_8(12).

  • Using the log property incorrectly, such as log⁡8(4)+log⁡8(12)=log⁡8(16)\log_8(4) + \log_8(12) = \log_8(16), is wrong. The correct rule is log⁡am+log⁡an=log⁡a(mn)\log_a m + \log_a n = \log_a(mn), so the sum is log⁡8(48)\log_8(48).

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