8-log-6-x-17-log-x-6-7-

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! \in {1, 2, 6, 24, 120, \dots} \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!