Whakaoti i te (1/16)^-3p=4^3

$$( \frac { 1 } { 16 } ) ^ { - 3 p } = 4 ^ { 3 }$$

Whakaoti mō p

$$p=\frac{1}{2}=0.5$$

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.

$$\left(\frac{1}{16}\right)^{-3p}=64$$

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

$$\log(\left(\frac{1}{16}\right)^{-3p})=\log(64)$$

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

$$-3p\log(\frac{1}{16})=\log(64)$$

Whakawehea ngā taha e rua ki te \log(\frac{1}{16}).

$$-3p=\frac{\log(64)}{\log(\frac{1}{16})}$$

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

$$-3p=\log_{\frac{1}{16}}\left(64\right)$$

Whakawehea ngā taha e rua ki te -3.

$$p=-\frac{\frac{3}{2}}{-3}$$