Multiple choice

If $a>1$ is a real number, then the number of roots of the equation $a^{2\log_2x}$ = 5+ 4${x}^{\log_2a}$ has

  1. 2

  2. infinite

  3. 0

  4. 1
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D Correct answer
Explanation

Let y = x^(log2 a). The equation becomes a^(2 log2 x) = y^2. Given a^(log2 x) = x^(log2 a), the equation simplifies to y^2 = 5 + 4y, or y^2 - 4y - 5 = 0. Factoring gives (y-5)(y+1) = 0, so y=5 or y=-1. Since y = x^(log2 a) must be positive, only y=5 is valid, yielding exactly one solution for x.

AI explanation

Using the properties of logarithms, we can rewrite the given equation a^(2 log2 x) = 5 + 4x^(log2 a) as x^(2 log2 a) - 4x^(log2 a) - 5 = 0. Letting y = x^(log2 a), the equation becomes a quadratic y^2 - 4y - 5 = 0, which factors as (y - 5)(y + 1) = 0. Since x is positive, y must be positive, leaving y = 5 as the only valid root. The equation x^(log2 a) = 5 has exactly one valid solution for any given base a > 1, so the number of roots is 1.

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