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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?

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

Commented by Tawa11 last updated on 23/Mar/25

Solve for x. I got x = 19

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

Commented by Ghisom last updated on 23/Mar/25

x=^? 19 (((19^(37) −19^(19) )/(19^(19) −19)))^(1/18) =19 (((19^(19) (19^(18) −1))/(19(19^(18) −1))))^(1/18) =19 (19^(18) )^(1/18) =19 true (((x^(3x−20) −x^x )/(x^x −x)))^(1/18) =(((x^x (x^(2x−20) −1))/(x(x^(x−1) −1))))^(1/18) = =(((x^(x−1) (x^(2x−20) −1))/(x^(x−1) −1)))^(1/18) we can only go on if 2x−20=x−1 ⇔ x=19

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

Commented by Tawa11 last updated on 23/Mar/25

(((x^(3x − 20) − x^x )/(x^x − x)))^(1/(x − 1)) = x ((x^(3x − 20) − x^x )/(x^x − x)) = x^(x − 1) x^(3x − 20) − x^x = x^(x − 1) (x^x − x) x^(3x − 20) − x^x = x^(2x − 1) − x^x x^(3x − 20) = x^(2x − 1) 3x − 20 = 2x − 1 x = 19

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

Commented by Tawa11 last updated on 23/Mar/25

19 is possible sir.

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

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