Whakaoti i te -8*4^x+1+18*4^x+1=10*4^-2

$$- 8 \times 4 ^ { x + 1 } + 18 \cdot 4 ^ { x + 1 } = 10 \times 4 ^ { - 2 }$$

Whakaoti mō x

$$x=-3$$

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

Pahekotia te -8\times 4^{x+1} me 18\times 4^{x+1}, ka 10\times 4^{x+1}.

$$10\times 4^{x+1}=10\times 4^{-2}$$

Tātaihia te 4 mā te pū o -2, kia riro ko \frac{1}{16}.

$$10\times 4^{x+1}=10\times \frac{1}{16}$$

Whakareatia te 10 ki te \frac{1}{16}, ka \frac{5}{8}.

$$10\times 4^{x+1}=\frac{5}{8}$$

Whakawehea ngā taha e rua ki te 10.

$$4^{x+1}=\frac{\frac{5}{8}}{10}$$

Tuhia te \frac{\frac{5}{8}}{10} hei hautanga kotahi.

$$4^{x+1}=\frac{5}{8\times 10}$$

Whakareatia te 8 ki te 10, ka 80.

$$4^{x+1}=\frac{5}{80}$$

Whakahekea te hautanga \frac{5}{80} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$4^{x+1}=\frac{1}{16}$$

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

$$\log(4^{x+1})=\log(\frac{1}{16})$$

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

$$\left(x+1\right)\log(4)=\log(\frac{1}{16})$$

Whakawehea ngā taha e rua ki te \log(4).

$$x+1=\frac{\log(\frac{1}{16})}{\log(4)}$$

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

$$x+1=\log_{4}\left(\frac{1}{16}\right)$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=-2-1$$

Whakaoti mō x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(2)}-3$$

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