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Question Number 146508 by mathdanisur last updated on 13/Jul/21
Find the sum of the roots of the equation: 16 ∙ 2^(∣x∣) + 5 ∙ 2^x = 42
$${Find}\:{the}\:{sum}\:{of}\:{the}\:{roots}\:{of}\:{the} \\ $$$${equation}: \\ $$$$\mathrm{16}\:\centerdot\:\mathrm{2}^{\mid\boldsymbol{{x}}\mid} \:+\:\mathrm{5}\:\centerdot\:\mathrm{2}^{\boldsymbol{{x}}} \:=\:\mathrm{42} \\ $$
Answered by Olaf_Thorendsen last updated on 13/Jul/21
16.2^(∣x∣) +5.2^x = 42 (1) 1) x = 0 16+5 ≠ 42 : impossible 2) x > 0 (1) : 16.2^x +5.2^x = 42 2^x = ((42)/(21)) = 2 ⇒ x = 1 3) x < 0 (1) : 16.2^(−x) +5.2^x = 42 5.2^(2x) −42.2^x +16 = 0 2^x = ((42±(√(42^2 −4×5×16)))/(10)) 2^x = ((42±38)/(10)) = (2/5) or 8 xln2 = ln2−ln5 or 3ln2 x = 1−((ln5)/(ln2)) (<0) or 3 (>0 : impossible) Finally, x = 1−((ln5)/(ln2)) Sum of the roots = 1+(1−((ln5)/(ln2))) = 2−((ln5)/(ln2))
$$\mathrm{16}.\mathrm{2}^{\mid{x}\mid} +\mathrm{5}.\mathrm{2}^{{x}} \:=\:\mathrm{42}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{x}\:=\:\mathrm{0} \\ $$$$\mathrm{16}+\mathrm{5}\:\neq\:\mathrm{42}\::\:\mathrm{impossible} \\ $$$$\left.\mathrm{2}\right)\:{x}\:>\:\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{16}.\mathrm{2}^{{x}} +\mathrm{5}.\mathrm{2}^{{x}} \:=\:\mathrm{42} \\ $$$$\mathrm{2}^{{x}} \:=\:\frac{\mathrm{42}}{\mathrm{21}}\:=\:\mathrm{2}\:\Rightarrow\:{x}\:=\:\mathrm{1} \\ $$$$\left.\mathrm{3}\right)\:{x}\:<\:\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{16}.\mathrm{2}^{−{x}} +\mathrm{5}.\mathrm{2}^{{x}} \:=\:\mathrm{42} \\ $$$$\mathrm{5}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{42}.\mathrm{2}^{{x}} +\mathrm{16}\:=\:\mathrm{0} \\ $$$$\mathrm{2}^{{x}} \:=\:\frac{\mathrm{42}\pm\sqrt{\mathrm{42}^{\mathrm{2}} −\mathrm{4}×\mathrm{5}×\mathrm{16}}}{\mathrm{10}} \\ $$$$\mathrm{2}^{{x}} \:=\:\frac{\mathrm{42}\pm\mathrm{38}}{\mathrm{10}}\:=\:\frac{\mathrm{2}}{\mathrm{5}}\:\mathrm{or}\:\mathrm{8} \\ $$$${x}\mathrm{ln2}\:=\:\mathrm{ln2}−\mathrm{ln5}\:\mathrm{or}\:\mathrm{3ln2} \\ $$$${x}\:=\:\mathrm{1}−\frac{\mathrm{ln5}}{\mathrm{ln2}}\:\left(<\mathrm{0}\right)\:\mathrm{or}\:\mathrm{3}\:\left(>\mathrm{0}\::\:\mathrm{impossible}\right) \\ $$$$\mathrm{Finally},\:{x}\:=\:\mathrm{1}−\frac{\mathrm{ln5}}{\mathrm{ln2}} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:=\:\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{ln5}}{\mathrm{ln2}}\right)\:=\:\mathrm{2}−\frac{\mathrm{ln5}}{\mathrm{ln2}} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$

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