Question-217937

Question Number 217937 by Tawa11 last updated on 23/Mar/25

Commented by Tawa11 last updated on 23/Mar/25

Is there any other possible answer ?

Commented by Tawa11 last updated on 23/Mar/25

Solve for x .

I got x = 19

Commented by Ghisom last updated on 23/Mar/25

x \overset{?}{=} 19

{(\frac{19^{37} - 19^{19}}{19^{19} - 19})}^{1 / 18} = 19

{(\frac{19^{19} (19^{18} - 1)}{19 (19^{18} - 1)})}^{1 / 18} = 19

{(19^{18})}^{1 / 18} = 19

true

{(\frac{x^{3 x - 20} - x^{x}}{x^{x} - x})}^{1 / 18} = {(\frac{x^{x} (x^{2 x - 20} - 1)}{x (x^{x - 1} - 1)})}^{1 / 18} =

= {(\frac{x^{x - 1} (x^{2 x - 20} - 1)}{x^{x - 1} - 1})}^{1 / 18}

we can only go on if

2 x - 20 = x - 1 \Leftrightarrow x = 19

Commented by Tawa11 last updated on 23/Mar/25

{(\frac{x^{3 x - 20} - x^{x}}{x^{x} - x})}^{\frac{1}{x - 1}} = x

\frac{x^{3 x - 20} - x^{x}}{x^{x} - x} = x^{x - 1}

x^{3 x - 20} - x^{x} = x^{x - 1} (x^{x} - x)

x^{3 x - 20} - x^{x} = x^{2 x - 1} - x^{x}

x^{3 x - 20} = x^{2 x - 1}

3 x - 20 = 2 x - 1

x = 19

Commented by Tawa11 last updated on 23/Mar/25

19 is possible sir .

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