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Question Number 200594 by cherokeesay last updated on 20/Nov/23

	Answered by Frix last updated on 21/Nov/23

	$Obviously x = 3 is a solution .$ $Then you must approximate . . . I found$ $x \approx . 211793616$ $x \approx - 2.80610974$
	Answered by Rasheed.Sindhi last updated on 21/Nov/23

	$3^{x - 1} . 5^{\frac{3 (x - 1)}{x}} = 3^{x^{2} - 7} . 5^{x^{2} - 7}$ $3^{x^{2} - 7 - x + 1} . 5^{x^{2} - 7 - \frac{3 (x - 1)}{x}} = 1$ $3^{x^{2} - x - 6} . 5^{\frac{x^{3} - 10 x + 3}{x}} = 1$ $3^{x^{2} - x - 6} = 5^{- \frac{x^{3} - 10 x + 3}{x}}$ $∙ x^{2} - x - 6 = 0 \land - \frac{x^{3} - 10 x + 3}{x} = 0$ $(x - 3) (x + 2) = 0 \land x^{3} - 10 x + 3 = 0$ $x = 3, - 2 \land (x - 3) (x^{2} + 3 x - 1) = 0$ $x = 3, - 2 \land x = 3, \frac{- 3 \pm \sqrt{13}}{2}$ $\Rightarrow x = 3$ $∙ (x^{2} - x - 6) \log 3 = (- \frac{x^{3} - 10 x + 3}{x}) \log 5$ $- \frac{x^{3} - 10 x + 3}{x (x^{2} - x - 6)} = \frac{\log 3}{\log 5} = \log_{5} 3$ $(\log_{5} 3 + 1) x^{3} - (\log_{5} 3) x^{2} - 3 ({2 \log}_{5} 3 + 5) x + 3 = 0$ $. . .$
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