1965 AHSME Problems/Problem 1

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Problem

The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:

$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 3 \qquad  \textbf{(E) }\ \text{more than 4}$


Solution 1

Solution by e_power_pi_times_i

Take the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$. Factoring results in $(2x-5)(x-1) = 0$, so there are $\boxed{\textbf{(C) } 2}$ real solutions.

Solution2

Notice that $a^0=1, a>0$. So $2^0=1$. So $2x^2-7x+5=0$. Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$. So this means that the equation has two real solutions. Therefore, select $\boxed{B}$.

~hastapasta

See Also

1965 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AHSME Problems and Solutions

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