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Sum the solutions of the equation \({2021^{2021x}} + {2021^{x + 1}}.x = {2021^{{x^2}}} + {2021^x}. {x^ 2}\).
A. \(2.\) B. \(2021.\) C. \(2022.\) D. \(2023.\)
The answer
Condition: \(x \in \mathbb{R}\)
Divide both sides of the equation by \({2021^x} > 0\), we get: \({2021^{2020x}} + 2021x = {2021^{{x^2} – x}} + {x ^2}\)
\( \Leftrightarrow {2021^{2020x}} + 2020x = {2021^{{x^2} – x}} + {x^2} – x\)
Consider the function \(f\left( t \right) = {2021^t} + t\) on \(\mathbb{R}\)
\(f’\left( t \right) = {2021^t}.\ln \left( {2021} \right) + 1 > 0,\,\forall t\)
\( \Rightarrow \) function \(f\left( t \right)\) covariates on \(\mathbb{R}\)
where \(f\left( {2021x} \right) = f\left( {{x^2} – x} \right)\)
From this it follows: \(2021x = {x^2} – x \Leftrightarrow \left[\begin{array}{l}x=0\\x=2022\end{array}\right\)[\begin{array}{l}x=0\x=2022\end{array}\right\)
So we have: \({x_1} + {x_2} = 2022\).
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These are VD-VDC sentences in the topic REVIEW OF CODE FUNCTIONS – LOGARIT.
