Whakaoti i te 2^x+1+2^x+1=20

$${ 2 }^{ x+1 } + { 2 }^{ x+1 } = 20$$

Whakaoti mō x

$$x=\log_{2}\left(5\right)\approx 2.321928095$$

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

Pahekotia te 2^{x+1} me 2^{x+1}, ka 2\times 2^{x+1}.

$$2\times 2^{x+1}=20$$

Whakawehea ngā taha e rua ki te 2.

$$2^{x+1}=\frac{20}{2}$$

Whakawehea te 20 ki te 2, kia riro ko 10.

$$2^{x+1}=10$$

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

$$\log(2^{x+1})=\log(10)$$

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

$$\left(x+1\right)\log(2)=\log(10)$$

Whakawehea ngā taha e rua ki te \log(2).

$$x+1=\frac{\log(10)}{\log(2)}$$

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

$$x+1=\log_{2}\left(10\right)$$

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

$$x=\log_{2}\left(10\right)-1$$

Whakaoti mō x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(2)}+\log_{2}\left(5\right)$$

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