These functions compute integer approximations to logarithms. They are based on the IEEE function logb.

For real values $x$ and positive numeric values $b$, the function ${\mathrm{ilog}}_b\left(x\right)$ computes the integer base $b$ logarithm of $\|x\|$; that is, the logarithm rounded down to the nearest integer.

If $b\&gt;1$, then ${\mathrm{ilog}}_b\left(x\right)$ returns $r$ such that $b^r\le \left|x\right|<b^{r\&plus;1}$.

If $b<1$, then ${\mathrm{ilog}}_b\left(x\right)$ returns $r$ such that $b^{r\&plus;1}<\left|x\right|\le b^r$.

For complex values $x$, ${\mathrm{ilog}}_b\left(x\right)$ computes $\max \left({\mathrm{ilog}}_b\left(\mathrm{\Re }\left(x\right)\right)\&comma;{\mathrm{ilog}}_b\left(\mathrm{\Im }\left(x\right)\right)\right)$.

The following relations hold for infinite and undefined values:

If $b\&gt;1$, then ${\mathrm{ilog}}_b\left(\pm \mathrm{\infty }\right)\&equals;\mathrm{\infty }$ and ${\mathrm{ilog}}_b\left(0\right)=-\mathrm{\infty }$.

If $b<1$, then ${\mathrm{ilog}}_b\left(\pm \mathrm{\infty }\right)\&equals;-\mathrm{\infty }$ and ${\mathrm{ilog}}_b\left(0\right)=\mathrm{\infty }$.

For all $b$, we have ${\mathrm{ilog}}_b\left(\mathrm{undefined}\right)=\mathrm{undefined}$.

You can enter the command ilog[b] using either the 1-D or 2-D calling sequence. For example, ilog[3](50) is equivalent to ${\mathrm{ilog}}_3\left(50\right)$.

The $\mathrm{ilog2}\left(x\right)$ and $\mathrm{ilog10}\left(x\right)$ functions compute the same values as ${\mathrm{ilog}}_2\left(x\right)$ and ${\mathrm{ilog}}_{10}\left(x\right)$, respectively. They both have more efficient implementations than the default algorithm for ${\mathrm{ilog}}_b\left(x\right)$ for $b\ne 2,b\ne 10$. (Indeed, ${\mathrm{ilog}}_2\left(x\right)$ and ${\mathrm{ilog}}_{10}\left(x\right)$ are generally computed using the $\mathrm{ilog2}$ and $\mathrm{ilog10}$ commands.)

There are sometimes situations where you need a rough approximation of the size of the number, where $\mathrm{ilog}$ to any base will do. Then using $\mathrm{ilog2}$ is probably the best choice. (It is even faster than $\mathrm{ilog10}$ for some cases.)

The $\mathrm{ilog}\left(x\right)$ function computes ${\mathrm{ilog}}_{\&ExponentialE;}\left(x\right)$, approximating the natural logarithm of $x$.

The computation of $\mathrm{ilog2}\left(x\right)$ and $\mathrm{ilog10}\left(x\right)$ is more efficient than ${\mathrm{ilog}}_b\left(x\right)$ for $b\ne 2\&comma;10$.

The ilog2 and ilog10 commands are thread-safe as of Maple 15.

For more information on thread safety, see index/threadsafe.

$\mathrm{ilog10}\left(x\right)$

$\mathrm{ilog10}\left(x\right)$

$\mathrm{ilog10}\left(150\right)$

$2$

$\mathrm{ilog10}\left({10}^{-37}\right)$

$\mathrm{-37}$

$\mathrm{ilog10}\left(2^{14}+3^{10}\mathrm{I}\right)$

$4$

$\mathrm{ilog2}\left(50\right)$

$5$

$\mathrm{ilog}\left[2\right]\left(2^8\right)$

$8$

$\mathrm{ilog}\left[3\right]\left(10\right)$

$2$

$\mathrm{ilog}\left(3\right)$

$1$

$\mathrm{\phi }≔\frac{1}{2}+\frac{1}{2}\mathrm{sqrt}\left(5\right)$

$\mathrm{\phi }≔\frac{1}{2}+\frac{\sqrt{5}}{2}$

$\mathrm{Lucas}≔n\mapsto 2\cdot \mathrm{combinat}:-\mathrm{fibonacci}\left(n+1\right)-\mathrm{combinat}:-\mathrm{fibonacci}\left(n\right)$

$\mathrm{Lucas}≔n\mapsto 2\cdot \mathrm{combinat}:-\mathrm{fibonacci}\left(n+1\right)-\mathrm{combinat}:-\mathrm{fibonacci}\left(n\right)$

$\mathrm{l2000}≔\mathrm{Lucas}\left(2000\right)$

$\mathrm{l2000}≔9446708185759308415384067495999677431530963218480368032804826598281856324445977322684945038267086094364761366000137291348836189673785457326607903364013465483957273836804336595888397782139002535468799414419546535346394066447256463745311310661259359973909189379826722425332112242554370313063917929424669185186291673823764654829513873821477637371237697744102254002802127905427315493403711022179894479121632130910668828127$

$\mathrm{evalf}\left[900\right]\left({\mathrm{\phi }}^{2000}\right)$

$9.44670818575930841538406749599967743153096321848036803280482659828185632444597732268494503826708609436476136600013729134883618967378545732660790336401346548395727383680433659588839778213900253546879941441954653534639406644725646374531131066125935997390918937982672242533211224255437031306391792942466918518629167382376465482951387382147763737123769774410225400280212790542731549340371102217989447912163213091066882812699999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999989414301994556404261178907219383372907698634419491617032649619167\times {10}^{417}$

$\mathrm{ilog}\left[\mathrm{\phi }\right]\left(\mathrm{Lucas}\left(2000\right)\right)$

$2000$

$\mathrm{l2001}≔\mathrm{Lucas}\left(2001\right)$

$\mathrm{l2001}≔15285094926360416404458847753853545672215113709866886761428113642426699623148315429402781518231396824903495901847342654693518581616001718736483909963503576057612526424919647262732272622204263764540041695502230453879379513883382816164600806848158718731671394389559139884843246782770607873339169023848390330701515748112231115788664367129870458859443365121656070623201452984720321258087284805794773049623826845661626456876$

$\mathrm{evalf}\left[900\right]\left({\mathrm{\phi }}^{2001}\right)$

$1.52850949263604164044588477538535456722151137098668867614281136424266996231483154294027815182313968249034959018473426546935185816160017187364839099635035760576125264249196472627322726222042637645400416955022304538793795138833828161646008068481587187316713943895591398848432467827706078733391690238483903307015157481122311157886643671298704588594433651216560706232014529847203212580872848057947730496238268456616264568760000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000654232116200611158245512923334354892212558579525250227369596382\times {10}^{418}$

$\mathrm{ilog}\left[\mathrm{\phi }\right]\left(\mathrm{Lucas}\left(2001\right)\right)$

$2000$