## 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})? – Math book

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. $$4$$.

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.