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Question Number 19135 by gourav~ last updated on 05/Aug/17

$$solveforx:$$$$2^{\mid x+2\mid }-\mid 2^{x+1}-1\mid =2^{x+1}+1$$

Answered by ajfour last updated on 05/Aug/17

$$\mathrm{x}+2⩾0\Rightarrow \mathrm{x}⩾-2$$$$2^{\mathrm{x}+1}-1⩾0\Rightarrow \mathrm{x}⩾-1$$$$\mathrm{Let}\mathrm{x}<-2$$$$\Rightarrow 2^{-\mathrm{x}-2}+2^{\mathrm{x}+1}-1=2^{\mathrm{x}+1}+1$$$$\mathrm{or}{2}^{-\mathrm{x}-2}=2$$$$\Rightarrow -\mathrm{x}-2=1$$$$\mathrm{x}=-3$$$$\mathrm{Let}-2⩽\mathrm{x}⩽-1$$$$\Rightarrow 2^{\mathrm{x}+2}+2^{\mathrm{x}+1}-1=2^{\mathrm{x}+1}+1$$$$\Rightarrow 2^{\mathrm{x}+2}=2$$$$\Rightarrow \mathrm{x}+2=1$$$$\mathrm{or}\mathrm{x}=-1$$$$\mathrm{Let}-1<\mathrm{x}$$$$\Rightarrow 2^{\mathrm{x}+2}-2^{\mathrm{x}+1}+1=2^{\mathrm{x}+1}+1$$$$\Rightarrow 2^{\mathrm{x}+2}=2\left(2^{\mathrm{x}+1}\right)$$$$\mathrm{True}\mathrm{for}\mathrm{all}\mathrm{x}>-1$$$$\Rightarrow \mathrm{x}\in [-1,\mathrm{\infty })\cup \{-3\}.$$

Commented by gourav~ last updated on 06/Aug/17

$$thankyousir$$

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