If log_b^2x+log_x^2b=1, where b>0, b≠1, and x≠1, then x equals:

1964 AMC 12 Problem 21

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21.

\log_{b^2}x+\log_{x^2}b=1,

where

b\gt0,

b\ne1,

and

x\ne1,

then

x

equals:

\dfrac1{b^2}

\dfrac1b

b^2

b

\sqrt b

Let

t=\log_bx

The equation becomes

\frac t2+\frac1{2t}=1

Let

t=\log_bx.

Then

\log_{b^2}x=\frac{t}{2}

and

\log_{x^2}b=\frac{1}{2t}.

Thus

t+\frac1t=2,

(t-1)^2=0

and

t=1.

Hence

x=b.

Therefore, the correct answer is D.