If x is positive and log x≥log2+12log x, then:

(E) the minimum value of x is 4

(A) x has no minimum or maximum value

(B) the maximum value of x is 1

(C) the minimum value of x is 1

(D) the maximum value of x is 4

1958 AMC 12 Problem 17

Attempt Problem 17 of the 1958 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1958 AMC 12 solutions, or check the answer key.

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17.

x

is positive and

\log x\ge\log2+\dfrac12\log x,

then:

x

has no minimum or maximum value

the maximum value of

x

1

the minimum value of

x

1

the maximum value of

x

4

the minimum value of

x

4

Subtract

\tfrac12\log x

from both sides

Double the resulting inequality and combine

2\log2

The inequality gives

\frac12\log x\ge\log2,

\log x\ge2\log2=\log4.

Hence

x\ge4,

and the minimum possible value of

x

4.

Therefore, the correct answer is E.