Whakaoti i te {left(1+3.2/100right)}^x=200

$${ \left(1+ \frac{ 3.2 }{ 100 } \right) }^{ x } =200$$

Whakaoti mō x

$$x=\log_{1.032}\left(200\right)\approx 168.207669123$$

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

Whakarohaina te \frac{3.2}{100} mā te whakarea i te taurunga me te tauraro ki te 10.

$$\left(1+\frac{32}{1000}\right)^{x}=200$$

Whakahekea te hautanga \frac{32}{1000} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$\left(1+\frac{4}{125}\right)^{x}=200$$

Tāpirihia te 1 ki te \frac{4}{125}, ka \frac{129}{125}.

$$\left(\frac{129}{125}\right)^{x}=200$$

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

$$\log(\left(\frac{129}{125}\right)^{x})=\log(200)$$

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

$$x\log(\frac{129}{125})=\log(200)$$

Whakawehea ngā taha e rua ki te \log(\frac{129}{125}).

$$x=\frac{\log(200)}{\log(\frac{129}{125})}$$

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

$$x=\log_{\frac{129}{125}}\left(200\right)$$

Whakaoti mō x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.032)}+\log_{1.032}\left(200\right)$$

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