2012 AIME I Problem 9

Question

Let

x,

y,

and

z

be positive real numbers that satisfy

\begin{aligned} 2\log_{x}(2y) &= 2\log_{2x}(4z) \\ &= \log_{2x^4}(8yz) \ne 0. \end{aligned}

The value of

xy^5z

can be expressed in the form

\frac{1}{2^{\frac{p}{q}}},

where

p

and

q

are relatively prime positive integers. Find

p + q.

Answer

Set

x = 2^a,

y = 2^b,

z = 2^c

so each logarithm becomes a ratio of linear expressions in the exponents

Equal ratios also equal their mediant:

\frac{u_1}{v_1} = \frac{u_2}{v_2} = \frac{u_1 + u_2}{v_1 + v_2}.

Compare that with the third ratio to pin down

a.

Write

x = 2^a,

y = 2^b,

z = 2^c.

Then

\log_x(2y) = \frac{b + 1}{a},

\log_{2x}(4z) = \frac{c + 2}{a + 1},

and

\log_{2x^4}(8yz) = \frac{b + c + 3}{4a + 1},

so the condition is

\begin{aligned} \frac{2(b + 1)}{a} &= \frac{2(c + 2)}{a + 1} \\ &= \frac{b + c + 3}{4a + 1} \ne 0. \end{aligned}

From the first two,

\frac{b + 1}{a} = \frac{c + 2}{a + 1},

and equal ratios also equal their mediant

\frac{b + c + 3}{2a + 1}.

Comparing with the third expression gives

\frac{2(b + c + 3)}{2a + 1} = \frac{b + c + 3}{4a + 1}.

The common value is nonzero, so

b + c + 3 \ne 0,

and thus

2(4a + 1) = 2a + 1,

giving

a = -\frac{1}{6}.

Then

\frac{b + 1}{-\frac{1}{6}} = \frac{c + 2}{\frac{5}{6}}

yields

c + 2 = -5(b + 1),

that is,

5b + c = -7.

Therefore

xy^5z = 2^{a + 5b + c} = 2^{-\frac{1}{6} - 7} = \frac{1}{2^{\frac{43}{6}}},

so

p + q = 43 + 6 = 49.

Problem 9 in other years