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Graham's number definition from: https://en.wikipedia.org/wiki/Graham%27s_number

Graham's number=G64

Of course the equation could be transformed in $\operatorname{G64}^x=2$, but I don't know how to deal with such great numbers.

PS: it is just a curiosity, it's not a question from ... real world..., so is purely theoretical, with no practical application, as far as I am concerned.

$x$ should be close to 0, but not 0.

vonbrand
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  • Well, whenever I calculate logarithms, in reality I calculate $$ \log_b a = \frac{\ln a}{\ln b} \quad \text{or} \quad \frac{\log a}{\log b} $$ Therefore, you just need to know the (natural) logarithm of the base. In this case, of the "large number". In the case of Graham's number, even the logarithm is a humongous number. – Matti P. Apr 15 '20 at 12:10
  • I can transform in every base I want, let say, base 2 or e, but... can it be calculated somehow? Base 2 it would be best... but, it will be hard to calculate log in base 2 of G64... and it would be 1/(log in base 2 of G64...) –  Apr 15 '20 at 12:14

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