Whakaoti i te {left(1+32/100right)}^x=200

$${ \left(1+ \frac{ 32 }{ 100 } \right) }^{ x } =200$$

Whakaoti mō x

$$x=\log_{\frac{33}{25}}\left(200\right)\approx 19.083975886$$

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

Whakahekea te hautanga \frac{32}{100} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$\left(1+\frac{8}{25}\right)^{x}=200$$

Tāpirihia te 1 ki te \frac{8}{25}, ka \frac{33}{25}.

$$\left(\frac{33}{25}\right)^{x}=200$$

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

$$\log(\left(\frac{33}{25}\right)^{x})=\log(200)$$

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

$$x\log(\frac{33}{25})=\log(200)$$

Whakawehea ngā taha e rua ki te \log(\frac{33}{25}).

$$x=\frac{\log(200)}{\log(\frac{33}{25})}$$

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

$$x=\log_{\frac{33}{25}}\left(200\right)$$

Whakaoti mō x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(\frac{33}{25})}+\log_{\frac{33}{25}}\left(200\right)$$

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