Whakaoti i te e^x+1=4

$$e ^ { x + 1 } = 4$$

Whakaoti mō x

$$x=2\ln(2)-1\approx 0.386294361$$

Ngā Upane e Whakamahi ana i te Tautuhinga o te Taupū Kōaro

Whakamahia ngā ture taupū me ngā taupū kōaro hei whakaoti i te whārite.

$$e^{x+1}=4$$

Tuhia te tau taupū kōaro o ngā taha e rua o te whārite.

$$\log(e^{x+1})=\log(4)$$

Ko te taupū kōaro o tētahi tau ka hīkina ki tētahi pū ko te pū whakarea ki te taupū kōaro o taua tau.

$$\left(x+1\right)\log(e)=\log(4)$$

Whakawehea ngā taha e rua ki te \log(e).

$$x+1=\frac{\log(4)}{\log(e)}$$

Mā te tikanga tātai huri pūtake \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$x+1=\log_{e}\left(4\right)$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=2\ln(2)-1$$

Whakaoti mō x (complex solution)

$$x=2\ln(2)-1+2\pi n_{1}i$$

$$n_{1}\in \mathrm{Z}$$