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:

Question

Accepted Answer

To solve the problem:

1. **First equation:**  
   From $$ \log_X (X^2 + 12) = 4 $$, use the logarithmic property:  
   $$
   \log_a b = c \implies a^c = b
   $$  
   Here, $$ X^4 = X^2 + 12 $$. Rearrange to form a quadratic equation:  
   $$
   X^4 - X^2 - 12 = 0
   $$  
   Substitute $$ Z = X^2 $$:  
   $$
   Z^2 - Z - 12 = 0
   $$  
   Factorize:  
   $$
   (Z - 4)(Z + 3) = 0
   $$  
   Since $$ Z > 0 $$, $$ Z = 4 $$. Thus, $$ X^2 = 4 $$, and $$ X = 2 $$ (as $$ X > 0 $$).

2. **Second equation:**  
   From $$ 3\log_Y X = 1 $$:  
   Divide by 3:  
   $$
   \log_Y X = \frac{1}{3}
   $$  
   Using the logarithmic property:  
   $$
   Y^{\frac{1}{3}} = X
   $$  
   Cube both sides:  
   $$
   Y = X^3
   $$  
   Substituting $$ X = 2 $$:  
   $$
   Y = 2^3 = 8
   $$

3. **Final calculation:**  
   Find $$ X + Y $$:  
   $$
   X + Y = 2 + 8 = 10
   $$

Thus, the correct answer is **10**.

Answer

Answer

Answer