1952 AMC 12 Problem 47

Question

In the set of equations

z^x=y^{2x},

2^z=2\cdot4^x,

x+y+z=16,

the integral roots in the order

x,

y,

z

are:

3,

4,

9

9,

-5,

12

12,

-5,

9

4,

3,

9

4,

9,

3

Compare exponents in

2^z=2\cdot4^x

to express

z

in terms of

x

For the intended positive solution,

z^x=y^{2x}

gives

z=y^2

; substitute both relations into the sum

For the intended positive integral solution, the first equation gives

z=y^2.

The second equation is

2^z=2^{2x+1},

so

z=2x+1

and

x=\frac{y^2-1}{2}.

Substituting into

x+y+z=16

gives

\frac{y^2-1}{2}+y+y^2=16.

The positive integral solution is

y=3,

which gives

x=4

and

z=9.

Direct substitution verifies all three displayed equations.

Thus, the intended listed answer is D.

Problem 47 in other years