Whakaoti i te 5^2x+1div125=25.f

$$5 ^ { 2 x + 1 } \div 125 = 25 . f$$

Whakaoti mō x

$$x=2$$

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.

$$\frac{1}{125}\times 5^{2x+1}=25$$

Me whakarea ngā taha e rua ki te 125.

$$5^{2x+1}=3125$$

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

$$\log(5^{2x+1})=\log(3125)$$

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

$$\left(2x+1\right)\log(5)=\log(3125)$$

Whakawehea ngā taha e rua ki te \log(5).

$$2x+1=\frac{\log(3125)}{\log(5)}$$

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

$$2x+1=\log_{5}\left(3125\right)$$

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

$$2x=5-1$$

Whakawehea ngā taha e rua ki te 2.

$$x=\frac{4}{2}$$

Whakaoti mō x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(5)}+2$$

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