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Question Number 26320 by byaw last updated on 24/Dec/17

solve x^2 −1=2^x

$$\mathrm{solve}\:{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{2}^{{x}} \\ $$

Commented by mrW1 last updated on 24/Dec/17

one solution is x=3.

$${one}\:{solution}\:{is}\:{x}=\mathrm{3}. \\ $$

Commented by jota@ last updated on 24/Dec/17

by graphing other solution is x≈−1.2

$${by}\:{graphing}\:{other}\:{solution}\:{is} \\ $$$${x}\approx−\mathrm{1}.\mathrm{2} \\ $$

Commented by mrW1 last updated on 25/Dec/17

there are totally three solutions: ≈−1.19825 =3 ≈3.40745

$${there}\:{are}\:{totally}\:{three}\:{solutions}: \\ $$$$\approx−\mathrm{1}.\mathrm{19825} \\ $$$$=\mathrm{3} \\ $$$$\approx\mathrm{3}.\mathrm{40745} \\ $$

Answered by ibraheem160 last updated on 24/Dec/17

((1±(√(17)))/2)

$$\frac{\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$

Commented by mrW1 last updated on 24/Dec/17

they don′t satisfy the equation.

$${they}\:{don}'{t}\:{satisfy}\:{the}\:{equation}. \\ $$

Answered by ibraheem160 last updated on 24/Dec/17

(1+1)^x =x^2 −1 1+x+((x(x−1).1^2 +....∞)/(2!))=x^2 −1 2+2x+x^2 −x=2x^2 −2 x^2 −x−4=0 x=((1±(√((−1)^2 −4(1×−4))))/(2×1)) x=((1±(√(17)))/2)

$$\left(\mathrm{1}+\mathrm{1}\right)^{{x}} ={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{1}+{x}+\frac{{x}\left({x}−\mathrm{1}\right).\mathrm{1}^{\mathrm{2}} +....\infty}{\mathrm{2}!}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{2}+\mathrm{2}{x}+{x}^{\mathrm{2}} −{x}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}×−\mathrm{4}\right)}}{\mathrm{2}×\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Dec/17

The bionomial expansion (1+y)^x =1+(x/1)y+((x(x−1))/(1.2))y^2 +((x(x−1)(x−2))/(1.2.3))y^3 +... +((x(x−1)...(x−r+1))/(1.2....r))x^r +... holds only when ∣y∣<1. You have used it for y=1

$$\mathrm{The}\:\mathrm{bionomial}\:\mathrm{expansion} \\ $$$$\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{x}} =\mathrm{1}+\frac{\mathrm{x}}{\mathrm{1}}\mathrm{y}+\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{2}}\mathrm{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\mathrm{y}^{\mathrm{3}} +... \\ $$$$\:\:\:\:\:+\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)...\left(\mathrm{x}−\mathrm{r}+\mathrm{1}\right)}{\mathrm{1}.\mathrm{2}....\mathrm{r}}\mathrm{x}^{\mathrm{r}} +... \\ $$$$\mathrm{holds}\:\mathrm{only}\:\mathrm{when}\:\mid\mathrm{y}\mid<\mathrm{1}. \\ $$$$\mathrm{You}\:\mathrm{have}\:\mathrm{used}\:\mathrm{it}\:\mathrm{for}\:\mathrm{y}=\mathrm{1} \\ $$

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