1965 AMC 12 Problem 1

Attempt Problem 1 of the 1965 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 1965 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

The number of real values of

x

satisfying the equation

2^{2x^2-7x+5}=1

is:

0

1

2

4

more than

4

A positive base other than

1

has value

1

only at exponent

0

Solve

2x^2-7x+5=0

Since the base is

2,

the exponent must be zero. Factoring gives

\begin{gathered} 2x^2-7x+5=0,\\ (2x-5)(x-1)=0. \end{gathered}

Thus

x=\frac{5}{2}

x=1,

giving two real values.

Therefore, the correct answer is C.

Problem 1 in other years