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Question Number 72344 by Rio Michael last updated on 27/Oct/19

$$provethat{e}^{lnx}=x$$$$or{a}^{{log}_{a}x}=x$$

Commented by Prithwish sen last updated on 27/Oct/19

$$\mathbf{let}\mathbf{lnx}=\mathbf{M}$$$$\therefore {\mathbf{e}}^{\mathbf{M}}=\mathbf{x}$$$$\mathbf{putting}\mathbf{the}\mathbf{value}\mathbf{of}\mathbf{M}$$$${\mathbf{e}}^{\mathbf{lnx}}=\mathbf{x}\mathbf{proved}$$

Commented by Tony Lin last updated on 27/Oct/19

$$let{log}_{a}x=y$$$$a^y=x$$$$\Rightarrow {log}_{a}x=y$$$$\Rightarrow a^{{log}_{a}x}=x$$

Commented by Rio Michael last updated on 30/Oct/19

$$thankyousirs$$

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