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Quite new here and after researching I haven't seen the thing I was looking for. Also, my apologies if I use wrong math variables, I am an enthusiast, not a professional.

The question is: For x= 1 to infinity and y = 1 to infinity, "in $2^x - 3^y = z$ , is $z$ increasing for any combination $x$ and $y$, while $z$ is positive ?" To further clarify below some examples.

Examples:

$2^5 - 3^3 = 5 $

$2^4 - 3^2 = 7$

$2^8 - 3^5 = 13$

...

$2^{152} - 3^{95} = 3.588096*10^{45}$

$2^{152} - 3^{96} = -6.536947*10^{44}$ negative and therefore not valid

$2^{153} - 3^{2} = 1.141798*10^{46}$

This continues at least for any combination for $x=450, y=283$. But is this for any value of $x$ and $y$? Or is there somewhere a difference that is stopping this trend and has a very tiny (positive) difference? For example: $2^{12345678} - 3^{123456} = 9.$

In other words, is there a proof that states that this trend is continueing?

A lot of thanks in advance.

EDIT:

There is a trend that for any $x$ combined with any $y$ the equation:

$2^x - 3^y = z$

where $z > 0$, $z$ will increase and never jump back to a (respectively) small difference.

Additional info since it is being flagged as unclear by some: Is there a $2^x$ slightly higher than $3^y$? (e.g. $100>z>1$ for a value $x > 8$: the lowest difference is $2^8-3^4=175$. OR $1000>z>1$ for a value $x > 11$: the lowest difference = $2^{11}-3^6=1319$) If there is a value $z$ for such a value $x$, that would make a big jump down.

@minus has prove the value $z$ will increase, see comment in answer below.

RobinvG
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2 Answers2

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It's not clear what you're asking, but (modifying MJD's suggestion in the comments) one possibility is that you want to know whether we can ever have all the following simultaneously:

  • $a<c$ and $b<d$ and $a-b<c-d$,
  • $2^a-3^b>0$ and $2^c-3^d>0$, but
  • $2^a-3^b>2^c-3^d$.

If so, this does happen. For example, $0<2^{149}-3^{94}<2^{146}-3^{92}$.

Especially Lime
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  • Thanks, but this is not what I am asking. I tried in the edit to more specific. The trend is that the difference in $z = 2^x - 3^y$ is getting bigger and bigger for $z$ is positve. I wanted to know if there is a break in the increasing value for $z$ so that $2^x$ is slightly bigger than $3^y$ – RobinvG Jun 03 '19 at 12:30
  • @RobinvG I don't understand what you mean. $z$ decreases when you go from $146,92$ to $149,94$, so doesn't that break the increasing trend? – Especially Lime Jun 03 '19 at 12:38
  • It does and I can see the confusion. My example of $2^5 -3^3$ is also lower than $2^4-3^2$ which already proves that. But if you arrange all the differences $z$ from small to big, you see a trend that the $z$ grows with the $x$. But does that mean there is no $2^x$ slightly (compared to the upward trend) higher than $3^y$? (e.g. $100 > z > 1$ for a value $x$ > 8 (lowest difference: 2^8-3^4=175) OR $1000 > z > 1$ for a value $x$ > 11(lowest difference: 2^11-3^6=1319) If there is, that would make a big jump down for $z$ for such a value $x$. ) – RobinvG Jun 03 '19 at 13:03
  • For R, this is what I made to see the trend and I hope it is self explanatory what I am doing here.

    df<-data.frame(1,2,3) names(df)[1] <- "x" names(df)[2] <- "y" names(df)[3] <- "z"

    for (x in 3:200) { for (y in 2:200){ if (((2^x)-(3^y))>0) { z <- 0 z <- (2^x)-(3^y) df<-rbind(df,c(x,y,z)) } } } df[order(df$z),]

     – RobinvG Jun 03 '19 at 13:15
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From what I gather, assuming we are working in the real numbers;

This is not true for all x and y.

$2^{x} - 3^{y} > 0$

$ xln(2) > yln(3)$

$x/y > ln(3)/ln(2)$

As shown by Minus one-twelth’s comment, you can also show that the further away from the ratio we get, the bigger the difference.

user679128
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  • Thanks. What you are saying here is that if $x > y*ln(3)/ln(2)$ you will get a power of 2 higher than a power of 3 right? Does it also mean that that the closer you get the x/y ratio to the $ln(3)/ln(2)$ ratio, the lower the difference is between $2^x -3^y$? – RobinvG Jun 03 '19 at 11:31
  • Yep that’s exactly what I was saying! Good question. I did not state anything to prove that, but I think you can find a counter example to show this is not true. I would think of this as a threshold, once you are over the threshold then everything is good. Hope that makes sense – user679128 Jun 03 '19 at 11:35
  • It makes sense what you are saying, but the ratio thing made no sense what I wrote ;) $2^5 - 3^3 = 5 (ratio: 5/3= 1,66667)$ and $2^{160} - 3^{102} = 1.207609*10^{48}$ (ratio: 160/102 = 1,569). Latter is closer to $ln(3)/ln(2)$, but difference is bigger – RobinvG Jun 03 '19 at 11:41
  • It helps me in a good direction, but it is unfortunately not an answer yet to proof that the difference for the equition can't all of a sudden drop for any $x$ or $y$. See the comment I wrote to MinusOne-Twelfth – RobinvG Jun 03 '19 at 11:44
  • $\newcommand{\e}{\varepsilon}$If $\frac{x}{y} = \frac{\ln 3}{\ln 2}+\e$, then $2^x = 3^y \times 2^{\e y}$, so $2^{x} - 3^{y} = \left(2^{\e y}-1\right)\times 3^{y}$. So if $y$ is big, you need $\e$ to be very small (i.e. the ratio to be very close to $\frac{\ln 3}{\ln 2}$) for this difference to be small. – Minus One-Twelfth Jun 03 '19 at 11:45
  • @RobinvG check minus one-twelfths comment, it shows that the further away you are the bigger the difference. – user679128 Jun 03 '19 at 11:51
  • I did, but first had to find out what epsilon means in math. Got it now and sorry for asking such a question, but not always understand the answer. If you say@MinusOne-Twelfth proves that the higher the numbers $x$ and $y$ are, the bigger difference I am grateful. Thank you very much and I try and study his answer a bit more. Thank you all. Check mark you asnwer as the answer is enough to close? Or should I also check mark Minus' answer somewhere? – RobinvG Jun 03 '19 at 12:04