Question

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

Accepted Answer

D. Given $\log_x (x^2 + 12) = 4 \implies x^4 = x^2 + 12$. Let $x^2 = t$. Then $t^2 - t - 12 = 0 \implies (t - 4)(t + 3) = 0$. Since $t = x^2 > 0$, we have $t = 4 \implies x = 2$ (since $x > 0$). Also, $3 \log_y x = 1 \implies \log_y x = \frac{1}{3} \implies y^{1/3} = 2 \implies y = 8$. Therefore, $x + y = 2 + 8 = 10$.

Logarithmic Equation Solution

Explanation

Practise this under exam conditions