Question

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

• A. 11
• B. 20
• C. 10
• D. 68

Answer 1

Answer: (C)

10

Answer 2

We have $\log_x(x^2 + 12) = 4$
$⇒ x ^2 +12=x ^4$
$⇒x^ 4 −x^ 2 −12=0$
$x^2(x^2 - 4) + 3(x^2 - 4) = 0$
$(x ^2 −4)(x^ 2 +3)=0$
given that x is a positive real number, then x = 2.

Given $3\log_y{x} = 1$
$\log_y{x} = \frac{1}{3}$
$⇒$ $x = y^\frac{1}{3}$
$⇒$ $y=x^ 3$
$⇒$ $y = 8.$
$⇒$ $x + y = 2 + 8 = 10.$