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Question Number 168383 by cortano1 last updated on 09/Apr/22

Commented by mr W last updated on 09/Apr/22

$$x=2^{192}$$

Answered by greogoury55 last updated on 09/Apr/22

$$\{\begin{array}{l}{t}^{256}=\log {}_t\left(x\right)\Rightarrow x=t^{t^{256}}\\ t^{128}=\log {}_t\left(\log {}_t\left(2\right)\right)+\log {}_t\left(24\right)-128\end{array}$$$$\Rightarrow t^{128}=\log {}_t\left(24\log {}_t\left(2\right)\right)-128$$$$\Rightarrow {[\log {}_t(24.\log {}_t\left(2\right))-128]}^2=\log {}_t\left(x\right)$$

Answered by mr W last updated on 10/Apr/22

$${\log }_t\left(24{\log }_{t}2\right)-128=t^{128}$$$${\log }_t\left({\log }_t{2}^{24}\right)=128+t^{128}$$$${\log }_t{2}^{24}=t^{128+t^{128}}=t^{128}{t}^{t^{128}}$$$$2^{24}=t^{t^{128}{t}^{t^{128}}}={\left(t^{t^{128}}\right)}^{t^{t^{128}}}$$$$leta=t^{t^{128}}$$$$\Rightarrow a^a=2^{24}=2^{3\times 8}=8^8\Rightarrow a=8$$$$or$$$$e^{\ln a}\ln a=24\ln 2$$$$\Rightarrow \ln a=W\left(24\ln 2\right)$$$$\Rightarrow a=e^{W\left(24\ln 2\right)}=\frac{24\ln 2}{W\left(24\ln 2\right)}=8$$$$t^{t^{128}}=a$$$${\left(t^{128}\right)}^{t^{128}}=a^{128}=8^{128}$$$$letb=t^{128}$$$$b^b=8^{2\times 64}={64}^{64}\Rightarrow b=64$$$$or$$$$\Rightarrow b=t^{128}=\frac{384\ln 2}{W\left(384\ln 2\right)}=64$$$$\Rightarrow t={64}^{\frac{1}{128}}=2^{\frac{3}{64}}$$$$$$$${\log }_{t}x=t^{256}$$$$\Rightarrow x=t^{t^{256}}=t^{t^{128}{t}^{128}}={\left(t^{t^{128}}\right)}^{t^{128}}=a^b=8^{64}=2^{192}$$

Commented by Tawa11 last updated on 09/Apr/22

$$\mathrm{Great}\mathrm{sir}.$$

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