Whakaoti i te 2^x+1=4

$$2 ^ { x + 1 } = 4$$

Whakaoti mō x

$$x=1$$

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.

$$2^{x+1}=4$$

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

$$\log(2^{x+1})=\log(4)$$

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

$$\left(x+1\right)\log(2)=\log(4)$$

Whakawehea ngā taha e rua ki te \log(2).

$$x+1=\frac{\log(4)}{\log(2)}$$

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

$$x+1=\log_{2}\left(4\right)$$

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

$$x=2-1$$

Whakaoti mō x (complex solution)

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

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