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Question Number 106630 by zahaku last updated on 06/Aug/20

$$Findninthisequation:$$$${(-2)}^n=4096$$

Answered by bemath last updated on 06/Aug/20

$${(-2)}^{\mathrm{n}}=\left(1024\right)\times 4={\left(2\right)}^{12}={(-2)}^{12}$$$$\to \mathrm{n}=12.@\mathrm{bemath}@$$

Commented by zahaku last updated on 06/Aug/20

$$There{isn}^{\prime }tanotherwaytofindn$$$$inthesameequation?$$

Answered by Aziztisffola last updated on 06/Aug/20

$$niseven\Rightarrow n=2p$$$${(-2)}^n=4096\iff {(-2)}^{2p}=4096$$$$\iff 4^{\mathrm{p}}=4096\Rightarrow \ln \left(4^{\mathrm{p}}\right)=\ln \left(4096\right)$$$$\ln \left(4\right)\mathrm{p}=\ln \left(4096\right)\Rightarrow \mathrm{p}=\frac{\ln \left(4096\right)}{2\ln \left(2\right)}$$$$2\mathrm{p}=\frac{\ln \left(4096\right)}{\ln \left(2\right)}=12$$$$2\mathrm{p}=12=\mathrm{n}\Rightarrow \mathrm{n}=12.$$

Commented by zahaku last updated on 06/Aug/20

$$Pleasecanyouexplainhow{(-2)}^{2p}becomes4p?$$

Commented by Aziztisffola last updated on 06/Aug/20

$${(-2)}^{2\mathrm{p}}={\left({(-2)}^2\right)}^{\mathrm{p}}=4^{\mathrm{p}}$$

Commented by zahaku last updated on 06/Aug/20

$$Thankyou.$$

Commented by Aziztisffola last updated on 06/Aug/20

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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