Question Number 186690 by mathlove last updated on 08/Feb/23
Answered by a.lgnaoui last updated on 08/Feb/23
$$\mathrm{x}×\mathrm{2}^{\mathrm{x}^{\mathrm{2}} } =\mathrm{4}\:\:\:\:\left(\mathrm{x}=\mathrm{a}^{\mathrm{2}} \right) \\ $$$$\mathrm{x}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{x}^{\mathrm{2}} } =\mathrm{4x} \\ $$$$\mathrm{posons}\:\mathrm{x}=\mathrm{z} \\ $$$$\mathrm{z}×\mathrm{2}^{\mathrm{z}} =\mathrm{2}^{\mathrm{2}} \mathrm{z} \\ $$$$\mathrm{logz}+\mathrm{zlog2}=\mathrm{2log2}+\mathrm{logz} \\ $$$$\mathrm{zlog2}=\mathrm{2log2} \\ $$$$\mathrm{z}=\mathrm{2}\:\:\:\:\: \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2}=\left(\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} =\sqrt{\mathrm{2}} \\ $$$${a}=^{\mathrm{4}} \sqrt{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 08/Feb/23
$${wrong}\:{again}! \\ $$$${many}\:{errors}! \\ $$$${a}^{{a}} \neq{a}^{\mathrm{2}} \\ $$$${a}^{{a}^{\mathrm{2}} } ={a}^{{a}×{a}} =\left({a}^{{a}} \right)^{{a}} \neq\left({a}^{{a}} \right)^{\mathrm{2}} \\ $$
Commented by aba last updated on 08/Feb/23
$$\mathrm{it}'\mathrm{s}\:\mathrm{correct} \\ $$
Commented by aba last updated on 08/Feb/23
$$\mathrm{he}\:\mathrm{posed}\:\mathrm{x}=\mathrm{a}^{\mathrm{a}} \\ $$
Commented by aba last updated on 08/Feb/23
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Commented by mr W last updated on 08/Feb/23
$${where}\:{did}\:{you}\:{study}\:{mathematics}\:{sir}? \\ $$$$\mathrm{2}^{{a}^{{a}^{\mathrm{2}} } } \neq\mathrm{2}^{\left({a}^{{a}} \right)^{\mathrm{2}} } \\ $$$${try}\:{this}: \\ $$$$\mathrm{3}^{\mathrm{3}^{\mathrm{2}} } =\mathrm{3}^{\mathrm{9}} =\mathrm{19683} \\ $$$$\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{27}^{\mathrm{2}} =\mathrm{729} \\ $$
Commented by aba last updated on 08/Feb/23
$$\mathrm{oh}\:\mathrm{ok}\:\mathrm{i}\:\mathrm{see}.\:\mathrm{You}\:\mathrm{are}\:\mathrm{right} \\ $$$$\mathrm{sorry} \\ $$
Commented by mr W last updated on 08/Feb/23
$${alright}\:{sir}! \\ $$