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Could not find a calculator online that could handle my large number. Could some help me with the solution for this very large number, I've forgotten how to divide exponentials with different bases.

$$\dfrac{2^{64^{64}}}{10^{80}}=?$$

Please show how you did it so I can do it myself too.

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  • $\begingroup$ @You'reInMyEye Sorry, it said I was not registered here at Math so posted using my email, then I connected it to my existing account but still says wrong user name. Don't know how to fix that now. $\endgroup$
    – temp88786
    Jul 16, 2016 at 14:12
  • $\begingroup$ This number is extremely big. $64^{64}$ contains about $115$ digits. If you try to raise $2$ to such a high exponent, you will get a number which has no possible practical use. $10^{80}$ is very small compared to $2^{64^{64}}$ $\endgroup$
    – Yuriy S
    Jul 16, 2016 at 14:23
  • $\begingroup$ What @You'reInMyEye said. Why do you care about that number? It is stupid big compared to anything sensible by which I mean it is many orders bigger then the number of atoms in the universe. $\endgroup$
    – DRF
    Jul 16, 2016 at 14:27
  • $\begingroup$ You used a different email address @IngeEivindHenriksen. Try to follow the help-centre recipe to merge accounts. $\endgroup$
    – Daniel Fischer
    Jul 16, 2016 at 14:28
  • $\begingroup$ @DanielFischer Thank you. $\endgroup$
    – temp88786
    Jul 16, 2016 at 14:30

3 Answers 3

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Dr. Graubner's answer is spot on but for a different approach that might be easier to use when you don't have a calculator (or are fairly bad with logs) you can take the following very rough approach.

$10^{80}<16^{80}=2^{4*80}=2^{320}$ so $$\frac{2^{64^{64}}}{10^{80}}>\frac{2^{64^{64}}}{2^{320}}=2^{64^{64}-320}$$

Now $64^{64}-320$ is so close to $64^{64}$ it's almost not worth bounding but if you really wanted to (and the error here is stupid large) you can notice that $64^{63}>320$ so $64^{63}*64-320>64^{63}*64-64^{63}=64^{63}*63>64^{63}$

Altogether then you get

$$\frac{2^{64^{64}}}{10^{80}}>2^{64^{63}}$$

The error is REALLY big but at least you get some idea without having to do any computations.

EDIT: I let Wolfram Alpha actually do the computations for me and the error you get my way is stupidly big.

$$2^{64^{64}}\approx 10^{10^{115.0741281073144}}$$ and ... wait for it $$\frac{2^{64^{64}}}{10^{80}}\approx 10^{10^{115.0741281073144}}$$

whereas $$2^{64^{63}}\approx 10^{10^{113.2679481333305}}$$

Further EDIT:

Thinking about this really should not have surprised me. Dividing by $10^{80}$ means throwing away $80$ zeros. The top has roughly $10^{115}$ zeros so ofcourse you can't possibly see the difference.

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  • $\begingroup$ Fantastic answer! $\endgroup$
    – temp88786
    Jul 16, 2016 at 15:32
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let $$x=\frac{2^{{64}^{64}}}{10^{80}}$$ then you will get $$\ln(x)=64^{64}\ln(2)-80\ln(10)$$

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Hint: $10 = 2\cdot 5$ so $10^{80} = 2^{80} \cdot 5^{80}$.

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