If \(x\) and \(y\) are positive real numbers such that \(log_x(x^2+12)=4\) and \(3\;log_yx=1\),then \(x+y\) equals

Question

If \( a, b, c \) are positive real numbers each distinct from unity, then the value of the determinant \[ \left| \begin{matrix} 1 & \log_a b & \log_a c \\ \log_b a & 1 & \log_b c \\ \log_c a & \log_c b & 1 \end{matrix} \right| \] is:

Answer 1

Given the equation \(\log_x(x^2 + 12) = 4\), where x is a positive real number, we derive \(x^2 + 12 = x^4\), which simplifies to \(x^4 - x^2 - 12 = 0\). Factoring this equation yields \((x^2 - 4)(x^2 + 3) = 0\). Since \(x^2 + 3\) cannot be zero for real x, we have \(x^2 - 4 = 0\), leading to \(x = 2\) (as x is positive).

Given \(3\log_y{x} = 1\), we get \(\log_y{x} = \frac{1}{3}\), which implies \(x = y^\frac{1}{3}\) and thus \(y = x^3\). Substituting \(x = 2\), we find \(y = 2^3 = 8\). Therefore, \(x + y = 2 + 8 = 10\).

If \(x\) and \(y\) are positive real numbers such that \(log_x(x^2+12)=4\) and \(3\;log_yx=1\),then \(x+y\) equals

The Correct Option is C

Solution and Explanation

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