Whakaoti i te 14.400=4000(1.025)^x

$$14.400 = 4000 ( 1.025 ) ^ { x }$$

Whakaoti mō x

$$x=2\log_{1.025}\left(0.06\right)\approx -227.874689765$$

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

Whakawehea ngā taha e rua ki te 4000.

$$\frac{14.4}{4000}=1.025^{x}$$

Whakarohaina te \frac{14.4}{4000} mā te whakarea i te taurunga me te tauraro ki te 10.

$$\frac{144}{40000}=1.025^{x}$$

Whakahekea te hautanga \frac{144}{40000} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 16.

$$\frac{9}{2500}=1.025^{x}$$

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

$$1.025^{x}=\frac{9}{2500}$$

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

$$\log(1.025^{x})=\log(\frac{9}{2500})$$

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

$$x\log(1.025)=\log(\frac{9}{2500})$$

Whakawehea ngā taha e rua ki te \log(1.025).

$$x=\frac{\log(\frac{9}{2500})}{\log(1.025)}$$

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

$$x=\log_{1.025}\left(\frac{9}{2500}\right)$$

Whakaoti mō x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.025)}+2\log_{1.025}\left(0.06\right)$$

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