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I want to solve the equation
$$2^a - 2^b = 2^x,$$

where $a$ and $b$ are extremely large numbers, for example $a =10^{10000 G}$, $B=10^{100G}$ with $G=10^{10^{100}}$, and I don't want to calculate the value of 2 to the power of $a$ or $b$ while solving the equation.

I hope this is the right place for my question, because I asked the same question here, but they said it is off-topic.

Maria
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    Are these meant to be integers? If so, then if $a>b$ we can rewrite the left hand to be $2^b(2^{a-b}-1)=2^x$ so we must have $b=x$. Or did you mean something else? – lulu Sep 28 '23 at 18:28
  • No, A can be a decimal number, and A is always greater than B. – Maria Sep 28 '23 at 18:32
  • You should edit to explain that, it's not obvious from what you wrote. What's wrong with the answers you got on the Mathematica site? – lulu Sep 28 '23 at 18:34
  • @lulu, it doesn't avoid the calculation 2^b or 2^(a-b) – Maria Sep 28 '23 at 18:35
  • @lulu,your answer is like that answers and it doesn't avoid the calculation 2^b or 2^(a-b) – Maria Sep 28 '23 at 18:37
  • @lulu,I now realize that using 2^A or 2^B is problematic for the performance I need to run on the computer, but 2^(a-b) is fine. – Maria Sep 28 '23 at 19:00
  • How is this going to work? Using the values of $a$ and $b$ in your question there are not enough hard-drives in the world to store $a$ or $b$ let alone do any calculations with them. – Peter Phipps Sep 28 '23 at 19:12
  • @Peter, you are right, I haven't tested with numbers that big and I'm currently trying with much smaller numbers like 2^100, but for a special case, I need to have the solution of the equation at least in theory. – Maria Sep 28 '23 at 19:25

2 Answers2

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Unless $a$ is close to $b$, $2^a>>2^b$ and the $2^b$ term can be dropped (and $x\approx a$) unless you need an exact answer.

In which case.

$2^a - 2^b = 2^b(2^{a-b} - 1) = 2^x\\ \log 2^b + \log (2^{a-b} - 1) = \log 2^x$

$b\log 2 + \log (2^{a-b} - 1) = x\log 2\\ x-b = \frac {\log (2^{a-b} - 1)}{\log 2}\\ x = \frac {\log(2^{a-b} - 1)}{\log 2} + b$

Log base 2 ($\lg$) might make this a little bit nicer as $\lg 2 = 1$

$x = \lg(2^{a-b} - 1) + b$

user317176
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Consider that you look for the zero of function $$f(x)=2^x-(2^a - 2^b) \qquad \text{with}\qquad a > b$$ Write it better as $$g(x)=x\log(2)-\log(2^a - 2^b)=x\log(2)-a\log(2)-\log(1-\epsilon)$$ with $\color{red}{\epsilon=2^{b-a}}$. This gives $$x=a+\frac{\log (1-\epsilon )}{\log (2)}\sim a-\frac{2^{b-a}}{\log (2)}$$

For $a=7$ and $b=5$, this gives $x=6.63933$ while the solution is $6.58496$.

Make it better using Taylor series which will write $$x=a-\frac{1}{\log (2)}\sum_{n=1}^\infty \frac {\epsilon^n} n$$

For the above example, using $p$ terms in the summation

$$\left( \begin{array}{cc} p & x_{(p)} \\ 1 & 6.63933 \\ 2 & 6.59424 \\ 3 & 6.58673 \\ 4 & 6.58532 \\ 5 & 6.58504 \\ \end{array} \right)$$

and you will need less and less terms to add when $(a-b)$ will be larger and larger.

You can even make it better if, instead of Taylor series, you use the $[n+1,n]$ Padé approximant $P_n$ of $\log (1-\epsilon )$.

For example $$P_2=-\frac{\epsilon \left(\epsilon ^2-21 \epsilon +30\right)}{9 \epsilon ^2-36 \epsilon +30}$$ whose error is $\frac{\epsilon ^6}{600}$ would give $x=6.58496$ (which is the solution).

Claude Leibovici
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