If b>1, x>0, and (2x)^log_b2-(3x)^log_b3=0, then x is

1981 AMC 12 Problem 15

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15.

b\gt1,

x\gt0,

and

(2x)^{\log_b2}-(3x)^{\log_b3}=0,

then

x

\frac1{216}

\frac16

1

6

not uniquely determined

Take logarithms to base

b

of the two equal positive powers

Let

A=\log_b2,

B=\log_b3,

and

u=\log_bx

Let

A=\log_b2,

B=\log_b3,

and

u=\log_bx.

Taking base-

b

logarithms gives

A(A+u)=B(B+u).

Since

A\ne B,

A+B+u=0.

Therefore

u=-\log_b6=\log_b(\frac{1}{6}),

x=\frac{1}{6}.

Therefore, the correct answer is B.