(√(8+log_6 (x!)))+(√(17−log


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Question Number 66413 by hmamarques1994@gmail.com last updated on 14/Aug/19

$\sqrt{8 + \log_{6} (x!)} + \sqrt{17 - \log_{x!} (6)} = 7$
	Answered by MJS last updated on 14/Aug/19
	$x!∈{1, 2, 6, 24, 120,...} ⇒ it′s easier to try ⇒ x=3 or put ln x! =t (√(8+(t/(ln 6))))+(√(17−((ln 6)/t)))=7 (√a)+(√b)=c a+2(√(ab))+b=c^2 2(√(ab))=c^2 −a−b 4ab=a^2 +b^2 +c^4 −2ac^2 −2ab−2bc^2 a^2 +b^2 +c^4 −2ac^2 −6ab−2bc^2 =0 ⇒ t^4 −116ln 6 t^3 +34ln^2 6 t^3 +80ln^3 6 t+ln^4 6 =0 (t−ln 6)(t^3 −115ln 6 t^2 −81ln^2 6 t−ln^3 6)=0 ⇒ t_1 =ln 6 ⇒ x_1 =3 the other solution give no values ∈N$
	$x! \in {1, 2, 6, 24, 120, . . .} \Rightarrow {it}^{'} s easier to try$ $\Rightarrow x = 3$ $or put \ln x! = t$ $\sqrt{8 + \frac{t}{\ln 6}} + \sqrt{17 - \frac{\ln 6}{t}} = 7$ $\sqrt{a} + \sqrt{b} = c$ $a + 2 \sqrt{a b} + b = c^{2}$ $2 \sqrt{a b} = c^{2} - a - b$ $4 a b = a^{2} + b^{2} + c^{4} - 2 {a c}^{2} - 2 a b - 2 {b c}^{2}$ $a^{2} + b^{2} + c^{4} - 2 {a c}^{2} - 6 a b - 2 {b c}^{2} = 0$ $\Rightarrow$ $t^{4} - 116 \ln 6 t^{3} + {34 \ln}^{2} 6 t^{3} + {80 \ln}^{3} 6 t + \ln^{4} 6 = 0$ $(t - \ln 6) (t^{3} - 115 \ln 6 t^{2} - {81 \ln}^{2} 6 t - \ln^{3} 6) = 0$ $\Rightarrow t_{1} = \ln 6 \Rightarrow x_{1} = 3$ $the other solution give no values \in N$
	Commented by hmamarques1994@gmai.com last updated on 14/Aug/19

	$G o o d!$
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