How to evaluate $a^b - b^a = x$

Question

I know the values of $a$ and $b$ and I want some other form of this equation to solve it using C++ programming language without overflow, because $a$ and $b$ can reach $100$.

I was thinking about logarithm but I could not complete the idea if we take the logarithm of the two sides:

$\log (a^b - b^a)$ = $\log x$

any idea?

MathJax hint: if you put a backslash before common functions, you get the right font and spacing. \log x gives $\log x$ in contrast to log x giving $log x$ — Ross Millikan, May 13 '18 at 14:49

C Monsour · Answer 1 · 2018-05-13T15:02:40.103

1

Evaluate $b \log(a)$ and $a \log(b)$. Suppose, for example, that $a \log(b)$ is smaller. $b \log(a) - a \log(b)=\log(\frac{a^b}{b^a})$. If this is so large that exponentiating causes overflow (or even gives you a 20-digit number), then for all practical purposes $x=a^b-b^a$ is just $a^b$ for purposes of the number of significant digits you have, and you are done (i.e., $\log(x)=b \log(a)$ to avoid overflow). If $b \log(a) - a \log(b)$ is less than $\log(10^{20})$ or so, exponentiate it and subtract $1$ to get $\frac{x}{b^a}$. If $b^a$ is not too large, solve directly. If $b^a$ is very large, take the log again to get $\log(x)-a\log(b)$, and then add $a\log(b)$ to get $\log(x)$.

The key thing is to realize that is $b \log(a)$ is sufficiently larger than $a \log(b)$, then $a^b-b^a$ will just be $a^b$. And of course (which I think you already know) to realize that if $x$ would overflow, you would still like to return $\log (x)$.

edited May 13 '18 at 15:02

answered May 13 '18 at 14:53

C Monsour

8,228

If the difference between $b\log(a)$ and $a\log(b)$ is greater than $\log (10)$ the larger is over $10$ times the smaller and the subtraction will not bring the computation into range. It is $b \log (a)$ that should be tested for being so large that exponentiating it will cause an overflow. – Ross Millikan May 13 '18 at 15:19
Your first statement would be true if you said greater than $10$, but as I said if exponentiating is large enough to cause overflow, then $a^b-b^a$ is functionally $a^b$ and the answer is basically $\log(x)$=$b\log(a)$. As for the second, I said if $b^a$ is not too large...that's called testing. I give a way to handle that case to get $\log(x)$. – C Monsour May 13 '18 at 15:46

Ross Millikan · Accepted Answer · 2018-05-13T15:05:30.020

0

Unless $a$ and $b$ are very close, one of these will be much larger than the other. In that case if either one overflows the difference will, too, so you might as well just do the straightforward computation. Contrast $$10^{12}=1,000,000,000,000\\12^{10}=61,917,364,224$$ Taking the difference doesn't reduce $10^{12}$ much at all. If $a,b$ are integers just test for equality. If they are equal return $0$, otherwise compute it as written and let it overflow.

edited May 13 '18 at 15:05

answered May 13 '18 at 14:48

Ross Millikan

374,822

I think the final answer should fit in long long, but the computations in between will overflow – AliOsm May 13 '18 at 14:53
I think the OP wants to get $\log(x)$ even if $x$ itself would overflow. I covered that in my answer. – C Monsour May 13 '18 at 14:54
You have a problem with your thousands separators. – ericw31415 May 13 '18 at 14:55
@AliOsm: my point is that will not happen. – Ross Millikan May 13 '18 at 15:05
@ericw31415: Thanks. Fixed. – Ross Millikan May 13 '18 at 15:05
@CMonsour: I didn't think so. I think OP thinks the subtraction will bring it back into range. Your point is a good one and I upvoted. – Ross Millikan May 13 '18 at 15:06
oh ok I got you, so there is no answer with other form, I think the only method is to use BigInteger class from Java – AliOsm May 13 '18 at 15:26

How to evaluate $a^b - b^a = x$

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