2×4^(x+2) −5×4^(x+1) −3×2^(2x+1) −4^x = 20


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Question Number 53877 by Mikael_Marshall last updated on 26/Jan/19

$2 \times 4^{x + 2} - 5 \times 4^{x + 1} - 3 \times 2^{2 x + 1} - 4^{x} = 20$
Commented by Abdo msup. last updated on 27/Jan/19

$l e t p u t 4^{x} = t \Rightarrow 32 t - 20 t - 6 t - t = 20 \Rightarrow$ $5 t = 20 \Rightarrow t = 4 a n d 4^{x} = t \Leftrightarrow e^{x l n (4)} = t \Rightarrow$ $x l n (4) = l n (t) \Rightarrow x = \frac{l n (t)}{l n (4)} b u t l n (t) = l n (4) \Rightarrow x = 1 .$
	Answered by byaw last updated on 26/Jan/19

	$16 \times 2 (2^{2 x}) - 5 \times 4 (2^{2 x}) - 3 \times 2 (2^{2 x}) - 2^{2 x} = 20$ $32 . 2^{2 x} - 20 . 2^{2 x} - 6 . 2^{2 x} - 1 . 2^{2 x} = 20$ $5 . 2^{2 x} = 5.4$ $2^{2 x} = 2^{2}$ $2 x = 2$ $x = 1$
	Commented by Mikael_Marshall last updated on 26/Jan/19

	$t h a n k s S i r . I^{'} m l e a r n i n g a l o t .$
	Answered by F_Nongue last updated on 26/Jan/19

	$2 \times 2^{2 x + 4} - 5 \times 2^{2 x + 2} - 3 \times 2^{2 x + 1} - 2^{2 x} = 20$ $2^{2 x + 5} - 5 \times 2^{2 x + 2} - 3 \times 2^{2 x + 1} - 2^{2 x} = 20$ $32 \times 2^{2 x} - 20 \times 2^{2 x} - 6 \times 2^{2 x} - 2^{2 x} = 20$ $2^{2 x} (32 - 20 - 6 - 1) = 20$ $2^{2 x} \times 5 = 20 / : 5$ $2^{2 x} = 4$ $2^{2 x} = 2^{2} \Rightarrow 2 x = 2 \Rightarrow x = 1$
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