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$$\large1+x^{\large10+\large\log x^{\large10+\large\log x^{10+\large ⋰}}}=\frac1{\log x}$$ To solve this equation I used $t=x^{\large10+\large\log x^{\large10+\large\log x^{10+\large ⋰}}}$ . LHS is equal to, $$1+t=1+x^{10+\log t}$$ $$t=x^{10+\log t}$$ $$\log t=(\log x)\times (10+\log t)$$ $$x=e^{\tfrac{\log t}{10+\log t}}$$ Hence the equation is equivalent to

$$t+1=\frac{10+\log t}{\log t}$$ $$t\log t=10$$

I'm not sure if this helps in solving the equation.

User
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1 Answers1

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You don't have to worry about the base, Let's just take the base to be $b$ then

$t\log_{b}(t)=10$

$\implies t\ln(t)=10\ln(b)$

$\implies W(\ln(t)e^{\ln(t)})=W(10\ln(b))$

$\implies\ln(t)=W(10\ln(b))$

$\implies t=e^{W(10\ln(b))}$

That makes,

$\textstyle\displaystyle{x=e^{\frac{W(10\ln(b))}{10+W(10\ln(b))}}}$

Rounak Sarkar
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