Whakaoti i te (1/25)quad=5^x+5

$$( \frac { 1 } { 25 } ) \quad = 5 ^ { x + 5 }$$

Whakaoti mō x

$$x=-7$$

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

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

$$\log(5^{x+5})=\log(\frac{1}{25})$$

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

$$\left(x+5\right)\log(5)=\log(\frac{1}{25})$$

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

$$x+5=\frac{\log(\frac{1}{25})}{\log(5)}$$

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

$$x+5=\log_{5}\left(\frac{1}{25}\right)$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$x=-2-5$$

Whakaoti mō x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(5)}-7$$

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