Whakaoti i te e^-3x+1=4

$${ e }^{ -3x+1 } =4$$

Whakaoti mō x

$$x=\frac{1-2\ln(2)}{3}\approx -0.128764787$$

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.

$$e^{-3x+1}=4$$

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

$$\log(e^{-3x+1})=\log(4)$$

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

$$\left(-3x+1\right)\log(e)=\log(4)$$

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

$$-3x+1=\frac{\log(4)}{\log(e)}$$

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

$$-3x+1=\log_{e}\left(4\right)$$

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

$$-3x=2\ln(2)-1$$

Whakawehea ngā taha e rua ki te -3.

$$x=\frac{2\ln(2)-1}{-3}$$

Whakaoti mō x (complex solution)

$$x=-\frac{i\times 2\pi n_{1}}{3}-\frac{2\ln(2)}{3}+\frac{1}{3}$$

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