If x, y>0, log_yx+log_xy=(10)/(3), and xy=144, then (x+y)/(2)=

1990 AMC 12 Problem 23

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23.

x,

y\gt0,

\log_yx+\log_xy=\frac{10}{3},

and

xy=144,

then

\frac{x+y}{2}=

12\sqrt2

13\sqrt3

24

30

36

Let

t=\log_yx,

so that

\log_xy=\frac{1}{t}

The resulting values of

t

show that one of

x,y

is the cube of the other

Let

t=\log_yx.

Then

t+\frac1t=\frac{10}{3},

3t^2-10t+3=0,

giving

t=3

\frac{1}{3}.

Thus one of

x,y

is the cube of the other. Let the smaller be

u.

Then

u^4=xy=144,

u=2\sqrt3

and

u^3=24\sqrt3.

Therefore

\frac{x+y}{2}=\frac{2\sqrt3+24\sqrt3}{2}=13\sqrt3.

Thus the correct answer is B.