adsense
How many pairs of integers \(\left( {x\,;y} \right)\) satisfy the conditions \(0 \le x \le 2020\) and \({\log _2}\left( {2x + 2} \right) + x – 3y = {8^y}\)?
A. \(2019\). B. \(2018\). C. \(1\). D. \(4\).
The answer
adsense
Since \(0 \le x \le 2020\) \({\log _2}\left( {2x + 2} \right)\) always makes sense.
We have \({\log _2}\left( {2x + 2} \right) + x – 3y = {8^y}\)
\( \Leftrightarrow {\log _2}\left( {x + 1} \right) + x + 1 = 3y + {2^{3y}}\)
\( \Leftrightarrow {\log _2}\left( {x + 1} \right) + {2^{{{\log }_2}\left( {x + 1} \right)}} = 3y + {2 ^{3y}}\) \(\left( 1 \right)\)
Consider the function \(f\left( t \right) = t + {2^t}\).
Defined set \(D = R\) and \(f’\left( t \right) = 1 + {2^t}\ln 2\)\( \Rightarrow \)\(f’\left( t \) right) > 0\), \(\forall t \in R\).
Infer that the function \(f\left( t \right)\) covariates on \(R\).
Thus \(\left( 1 \right) \Leftrightarrow {\log _2}\left( {x + 1} \right) = 3y\)\( \Leftrightarrow x + 1 = {2^{3y}}\) \( \Leftrightarrow y = {\log _8}\left( {x + 1} \right)\).
We have \(0 \le x \le 2020\) so \(1 \le x + 1 \le 2021\) deduces \(0 \le {\log _8}\left( {x + 1} \right) \le {\log _8}2021\).
Again there are \({\log _8}2021 \approx 3.66\) and \(y \in Z\) so \(y \in \left\{ {0\,;1\,;2\,;\ left. 3 \right\}} \right.\).
So there are 4 pairs of numbers \(\left( {x\,;y} \right)\) that fully satisfy the problem requirements are the pairs \(\left( {0\,;0} \right)\), \ (\left( {7\,;1} \right)\) ,\(\left( {63\,;2} \right)\),\(\left( {511\,;3} \right) \).
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These are VD-VDC sentences in the topic REVIEW OF CODE FUNCTIONS – LOGARIT.
