Whakaoti i te 2700=1800e^.19h

$$2700 = 1800 e ^ { .19 h }$$

Whakaoti mō h

$$h=\frac{100\ln(3)-100\ln(2)}{19}\approx 2.134026885$$

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

Whakawehea ngā taha e rua ki te 1800.

$$\frac{2700}{1800}=e^{0.19h}$$

Whakahekea te hautanga \frac{2700}{1800} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 900.

$$\frac{3}{2}=e^{0.19h}$$

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

$$e^{0.19h}=\frac{3}{2}$$

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

$$\log(e^{0.19h})=\log(\frac{3}{2})$$

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

$$0.19h\log(e)=\log(\frac{3}{2})$$

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

$$0.19h=\frac{\log(\frac{3}{2})}{\log(e)}$$

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

$$0.19h=\log_{e}\left(\frac{3}{2}\right)$$

Whakawehea ngā taha e rua o te whārite ki te 0.19, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

$$h=\frac{\ln(\frac{3}{2})}{0.19}$$

Whakaoti mō h (complex solution)

$$h=\frac{i\times 200\pi n_{1}}{19}+\frac{100\ln(3)}{19}-\frac{100\ln(2)}{19}$$

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