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Question Number 213923 by luj last updated on 21/Nov/24

Answered by MATHEMATICSAM last updated on 21/Nov/24

x^(32) = 2^x or 32ln∣x∣ = xln2 For x > 0 or ((lnx)/x) = ((ln2)/(32)) or x^(− 1) .lnx = ((ln2)/(32)) or lnx.e^(−lnx) = ((ln2)/(32)) or −lnx.e^(−lnx) = −((ln2)/(32)) or − lnx = W(− ((ln2)/(32))) or x = e^(− W(−((ln2)/(32)))) ((lnx)/x) = ((ln2)/(32)) or ((lnx)/x) = ((ln2^8 )/(8 × 32)) = ((ln256)/(256)) or x = 256 For x < 0 x = e^(−W(((ln2)/(32))))

$${x}^{\mathrm{32}} \:=\:\mathrm{2}^{{x}} \\ $$$${or}\:\mathrm{32ln}\mid{x}\mid\:=\:{x}\mathrm{ln2} \\ $$$$\mathrm{For}\:{x}\:>\:\mathrm{0} \\ $$$${or}\:\frac{\mathrm{ln}{x}}{{x}}\:=\:\frac{\mathrm{ln2}}{\mathrm{32}}\: \\ $$$${or}\:{x}^{−\:\mathrm{1}} .\mathrm{ln}{x}\:=\:\frac{\mathrm{ln2}}{\mathrm{32}} \\ $$$${or}\:\mathrm{ln}{x}.{e}^{−\mathrm{ln}{x}} \:=\:\frac{\mathrm{ln2}}{\mathrm{32}} \\ $$$${or}\:−\mathrm{ln}{x}.{e}^{−\mathrm{ln}{x}} \:=\:−\frac{\mathrm{ln2}}{\mathrm{32}} \\ $$$${or}\:−\:\mathrm{ln}{x}\:=\:{W}\left(−\:\frac{\mathrm{ln2}}{\mathrm{32}}\right) \\ $$$${or}\:{x}\:=\:{e}^{−\:{W}\left(−\frac{\mathrm{ln2}}{\mathrm{32}}\right)} \\ $$$$\frac{\mathrm{ln}{x}}{{x}}\:=\:\frac{\mathrm{ln2}}{\mathrm{32}} \\ $$$${or}\:\frac{\mathrm{ln}{x}}{{x}}\:=\:\frac{\mathrm{ln2}^{\mathrm{8}} }{\mathrm{8}\:×\:\mathrm{32}}\:=\:\frac{\mathrm{ln256}}{\mathrm{256}} \\ $$$${or}\:{x}\:=\:\mathrm{256} \\ $$$$\mathrm{For}\:{x}\:<\:\mathrm{0}\:\: \\ $$$${x}\:=\:{e}^{−{W}\left(\frac{\mathrm{ln2}}{\mathrm{32}}\right)} \\ $$

Answered by mr W last updated on 21/Nov/24

x^(32) =2^x x=±2^(x/(32)) =±e^((x/(32))ln 2) xe^(−(x/(32))ln 2) =±1 −((xln 2)/(32))e^(−(x/(32))ln 2) =±((ln 2)/(32)) −((xln 2)/(32))=W(±((ln 2)/(32))) ⇒x=−((32)/(ln 2))W(±((ln 2)/(32))) = { ((−((32)/(ln 2))W(((ln 2)/(32)))≈−0.97901693)),((−((32)/(ln 2))W(−((ln 2)/(32)))= { ((=256)),((≈1.02239294)) :})) :}

$${x}^{\mathrm{32}} =\mathrm{2}^{{x}} \\ $$$${x}=\pm\mathrm{2}^{\frac{{x}}{\mathrm{32}}} =\pm{e}^{\frac{{x}}{\mathrm{32}}\mathrm{ln}\:\mathrm{2}} \\ $$$${xe}^{−\frac{{x}}{\mathrm{32}}\mathrm{ln}\:\mathrm{2}} =\pm\mathrm{1} \\ $$$$−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}{e}^{−\frac{{x}}{\mathrm{32}}\mathrm{ln}\:\mathrm{2}} =\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}} \\ $$$$−\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}=\mathbb{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}\mathbb{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right) \\ $$$$\:\:=\begin{cases}{−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}\mathbb{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right)\approx−\mathrm{0}.\mathrm{97901693}}\\{−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}\mathbb{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right)=\begin{cases}{=\mathrm{256}}\\{\approx\mathrm{1}.\mathrm{02239294}}\end{cases}}\end{cases} \\ $$

Answered by kapoorshah last updated on 21/Nov/24

x^(1/x) = 2^(1/(32)) x^(1/x) = 2^((8/8) (1/(32))) x^(1/x) = 2^(8 ((1/8) (1/(32)))) x^(1/x) = 256^(1/(256)) x = 256

$${x}^{\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{32}}} \\ $$$${x}^{\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{2}^{\frac{\mathrm{8}}{\mathrm{8}}\:\frac{\mathrm{1}}{\mathrm{32}}} \\ $$$${x}^{\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{2}^{\mathrm{8}\:\left(\frac{\mathrm{1}}{\mathrm{8}}\:\frac{\mathrm{1}}{\mathrm{32}}\right)} \:\: \\ $$$${x}^{\frac{\mathrm{1}}{{x}}} \:=\:\mathrm{256}^{\frac{\mathrm{1}}{\mathrm{256}}} \\ $$$${x}\:=\:\mathrm{256} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 21/Nov/24

there are totally three real solutions.

$${there}\:{are}\:{totally}\:{three}\:{real}\:{solutions}. \\ $$

Commented by MATHEMATICSAM last updated on 21/Nov/24

yes

$$\mathrm{yes} \\ $$

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