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I am looking for formulas similar to classical logarithmic identities: $$a^{\log_ab}=b$$ $$a^{\log_cb}=b^{\log_ca}$$

involving floor or ceiling functions.

I need to evaluate the following expressions:

$$a^{\lfloor \log_ab \rfloor}$$ $$a^{\lceil \log_ab \rceil}$$

Is it possible to assert something about the last two expressions?

R. S.
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  • could invoke monotonicity of exponential function, can't really think of anything else – AlvinL Apr 26 '23 at 13:18
  • Interesting to make a contour plot https://www.wolframalpha.com/input?i=ContourPlot%5Ba%5EFloor%5BLog%5Bb%5D%2FLog%5Ba%5D%5D%2C+%7Ba%2C+1%2C+1000%7D%2C+%7Bb%2C+1%2C+1000%7D%5D – Claude Leibovici Apr 26 '23 at 13:26
  • Assuming that $~b \geq a,~$ consider placing $~b~$ somewhere between two of the terms of the sequence $~a^1, ~a^2, ~a^3, ~a^4, ~\cdots.$ – user2661923 Apr 26 '23 at 13:46

1 Answers1

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As the comment say the inequalities based on monotonicity seems the best we can conclude. We have $$\log_ab-1<\lfloor\log_ab\rfloor\le\log_ab,$$ and since the function $x\mapsto\log_ax$ is continuous this inequality cannot be sharpened. Indeed if $b=a^n$ for some integer $n$ we have the equality on the right, and $b$ can be arbitrarily close to some integer power of $a$ from the left so that $\lfloor\log_ab\rfloor$ is arbitrarily close to the lower bound. Again since the function $x\mapsto a^x$ is continuous the best we can conclude is $$b/a<\lfloor\log_ab\rfloor\le b$$ (if $a>1$) or $$b\le\lfloor\log_ab\rfloor<b/a$$ (if $0<a<1$).

Noiril
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