Multiple choice

The number of real roots of the equation $5+\left| { 2 }^{ x }-1 \right| ={ 2 }^{ x }\left( { 2 }^{ x }-2 \right) $ is:

  1. $2$
  2. $3$
  3. $4$
  4. $1$
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D Correct answer
Explanation

Let y = 2^x. Equation: 5 + |y - 1| = y(y - 2) = y^2 - 2y. Case 1: y >= 1: 5 + y - 1 = y^2 - 2y => y^2 - 3y - 4 = 0 => (y-4)(y+1) = 0. y = 4 (since y > 0). 2^x = 4 => x = 2. Case 2: y < 1: 5 - y + 1 = y^2 - 2y => y^2 - y - 6 = 0 => (y-3)(y+2) = 0. No solution for y < 1. Only 1 real root.

AI explanation

To remove the absolute value, assume the base condition $2^x - 1 \ge 0$, which means $2^x \ge 1$ and $x \ge 0$. Substituting this into the equation yields $5 + 2^x - 1 = 2^x(2^x - 2)$, which simplifies to $(2^x)^2 - 3(2^x) - 4 = 0$. Factoring this quadratic in terms of $2^x$ gives $(2^x - 4)(2^x + 1) = 0$, leading to $2^x = 4$ or $2^x = -1$. Since $2^x$ must be positive, $x = 2$ is the only valid root, and it satisfies our initial assumption of $x \ge 0$. The equation has exactly 1 real root.

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