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Difference between revisions of "1965 AHSME Problems/Problem 1"

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

Revision as of 11:35, 22 November 2016

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

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.

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