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Question Number 121175 by Ar Brandon last updated on 05/Nov/20

Answered by MJS_new last updated on 05/Nov/20

$$x<1$$$$g\left(x\right)<1$$

Commented by Ar Brandon last updated on 05/Nov/20

Any calculations or explanations, Sir ?��

Commented by MJS_new last updated on 05/Nov/20

$${\mathrm{e}}^x+{\mathrm{e}}^{g\left(x\right)}=\mathrm{e}\iff g\left(x\right)=\ln (\mathrm{e}-{\mathrm{e}}^x)$$$$\mathrm{e}-{\mathrm{e}}^x>0\Rightarrow \mathrm{e}>{\mathrm{e}}^x\Rightarrow x<1$$$$\underset{x\to 1^{-}}{\lim }g\left(x\right)=-\mathrm{\infty }$$$$\underset{x\to -\mathrm{\infty }}{\lim }g\left(x\right)=1$$$$\Rightarrow g\left(x\right)<1$$

Commented by Ar Brandon last updated on 05/Nov/20

It's clear now. Thanks Sir

Commented by MJS_new last updated on 05/Nov/20

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

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