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

Commented by Tawa11 last updated on 23/Mar/25

$$\mathrm{Is}\mathrm{there}\mathrm{any}\mathrm{other}\mathrm{possible}\mathrm{answer}?$$

Commented by Tawa11 last updated on 23/Mar/25

$$\mathrm{Solve}\mathrm{for}\mathrm{x}.$$$$\mathrm{I}\mathrm{got}\mathrm{x}=19$$

Commented by Ghisom last updated on 23/Mar/25

$$x\stackrel{?}{=}19$$$${\left(\frac{{19}^{37}-{19}^{19}}{{19}^{19}-19}\right)}^{1/18}=19$$$${\left(\frac{{19}^{19}({19}^{18}-1)}{19({19}^{18}-1)}\right)}^{1/18}=19$$$${\left({19}^{18}\right)}^{1/18}=19$$$$\mathrm{true}$$$$$$$${\left(\frac{x^{3x-20}-x^x}{x^x-x}\right)}^{1/18}={\left(\frac{x^x(x^{2x-20}-1)}{x(x^{x-1}-1)}\right)}^{1/18}=$$$$={\left(\frac{x^{x-1}(x^{2x-20}-1)}{x^{x-1}-1}\right)}^{1/18}$$$$\mathrm{we}\mathrm{can}\mathrm{only}\mathrm{go}\mathrm{on}\mathrm{if}$$$$2x-20=x-1\iff x=19$$

Commented by Tawa11 last updated on 23/Mar/25

$${\left(\frac{{\mathrm{x}}^{3\mathrm{x}-20}-{\mathrm{x}}^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{x}}-\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}-1}}=\mathrm{x}$$$$\frac{{\mathrm{x}}^{3\mathrm{x}-20}-{\mathrm{x}}^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{x}}-\mathrm{x}}={\mathrm{x}}^{\mathrm{x}-1}$$$${\mathrm{x}}^{3\mathrm{x}-20}-{\mathrm{x}}^{\mathrm{x}}={\mathrm{x}}^{\mathrm{x}-1}({\mathrm{x}}^{\mathrm{x}}-\mathrm{x})$$$${\mathrm{x}}^{3\mathrm{x}-20}-{\mathrm{x}}^{\mathrm{x}}={\mathrm{x}}^{2\mathrm{x}-1}-{\mathrm{x}}^{\mathrm{x}}$$$${\mathrm{x}}^{3\mathrm{x}-20}={\mathrm{x}}^{2\mathrm{x}-1}$$$$3\mathrm{x}-20=2\mathrm{x}-1$$$$\mathrm{x}=19$$

Commented by Tawa11 last updated on 23/Mar/25

$$19\mathrm{is}\mathrm{possible}\mathrm{sir}.$$

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