Whakaoti i te e^b+q+2=30

$$e ^ { b + q } + 2 = 30$$

Whakaoti mō b

$$b=-q+\ln(28)$$

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

Whakamahia ngā ture taupū me ngā taupū kōaro hei whakaoti i te whārite.

$$e^{b+q}+2=30$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$e^{b+q}=28$$

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

$$\log(e^{b+q})=\log(28)$$

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

$$\left(b+q\right)\log(e)=\log(28)$$

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

$$b+q=\frac{\log(28)}{\log(e)}$$

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

$$b+q=\log_{e}\left(28\right)$$

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

$$b=\ln(28)-q$$

Whakaoti mō q

$$q=-b+\ln(28)$$

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

Whakamahia ngā ture taupū me ngā taupū kōaro hei whakaoti i te whārite.

$$e^{q+b}+2=30$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$e^{q+b}=28$$

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

$$\log(e^{q+b})=\log(28)$$

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

$$\left(q+b\right)\log(e)=\log(28)$$

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

$$q+b=\frac{\log(28)}{\log(e)}$$

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

$$q+b=\log_{e}\left(28\right)$$

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

$$q=\ln(28)-b$$