Why can't logarithms be negative?

The only numbers you can plug into a logarithm are positive numbers. Negative numbers, and the number 0, aren’t acceptable arguments to plug into a logarithm, but why?
The reason has more to do with the base of the logarithm than with the argument of the logarithm. To understand why, we have to understand that logarithms are actually exponents. The base of a logarithm is also the base of a power function.
When you have a power function with base 0, the result of that power function is always going to be 0. In other words, there’s no exponent you can put on 0 that won’t give you back a value of 0. Or, put a different way, 0 raised to anything is always still 0. In the same way, 1 raised to anything is always still 1.
If you raise a negative number to a positive number that’s not an integer, but instead a fraction or a decimal, you might end up with a negative number underneath a square root. And as you know, unless we’re getting into imaginary numbers, we can’t deal with a negative number underneath a square root.
So 0, 1 and every negative number presents a potential problem as the base of a power function. And if those numbers can’t reliably be the base of a power function, then they also can’t reliably be the base of a logarithm.
For that reason, we only allow positive numbers other than 1 as the base of the logarithm. Then what we know is that, if the base of our power function is positive, it doesn’t matter what exponent we put on that base (it could be a positive number, a negative number, of 0), that power function is going to come out as a positive number.
So in summary, because the base can only be a positive number, that means the argument of the logarithm can only be a positive number. Which means that in order to protect our bases, we have to only allow positive arguments inside the logarithm.
0:00 // The argument can’t be negative
0:19 // Parts of the logarithm
0:30 // The argument of the logarithm can’t be negative because of how the base of the logarithm is defined
0:47 // The logarithm is a power function
1:36 // What kind of numbers can the base of the logarithm actually be?
3:11 // How does the base of the logarithm effect the argument of the logarithm?
4:32 // Summary and conclusion
Music by Joakim Karud: / joakimkarud
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[@kristakingmath]

Just to clarify, the argument of the logarithm can be 1. In fact, it can be any positive number (just not a negative number or 0). The base of the logarithm is the number that can't be 0, 1, or any negative number. The base of the logarithm can only be a positive number other than 1. If the base of the logarithm is negative, it means that the argument of the logarithm, or the logarithm in general, will be complex, meaning that it'll include the imaginary number i.

[@CE113378]

The argument of the logarithm can be negative. Say, L = ln (-1). Then, e^L = -1. Let L = j*x, where j = sqrt(-1). By Euler's identity, e^(jx) = cos (x) + j sin(x). So, e^jx = -1 = cos(x) + j sin(x). x = pi radians. L = j*pi. Arguably, in the limit, the argument of the logarithm can also approach 0. As the argument approaches 0, the logarithm approaches negative infinity. The base of a logarithm can also be negative, but that would imply that either the argument of the logarithm, or the logarithm itself, would be complex (possibly both). Cheers! I love your channel, BTW.

[@forthesakeofscience961]

2:59 are you sure.......?

[@shoutz5872]

@@CE113378 having a limit does not mean equal. also you forgot to include the multiples by k element of N multiple

[@jeremy.N]

great Video, but have you ever Heard of complex number?

[@OperationPitbull]

@@forthesakeofscience961 It is implied that she's only using real numbers for the basis of this video.

[@BlueHawkPictures17]

Well, you can have a negative argument as long as you have a complex base.

[@kristakingmath]

Yes, but if you're not going to introduce complex numbers... :)

[@BlueHawkPictures17]

What's your plan for future videos? Will it be on more advanced fields of math, focus on high school and college level math, or is there no particular plan in mind?

[@kristakingmath]

Yes, I'm hoping to add more videos for Differential Equations and Linear Algebra, plus Probability and Statistics. And I'd like to add other kinds of math courses for more "real-world" topics. :)

[@josezambrano5224]

Real world topics sound exciting. Looking forward to it.

[@mike4ty4]

If you are willing to admit complex numbers, you can have a negative argument with any base, including a positive real. The trouble is that when you go up to complex numbers, logarithms are no longer unambiguous: adding any integer multiple of 2pi i/log(base) (or i tau/log(base) ) to the logarithm gives another logarithm.

[@JohnnyYenn]

I've always thought that there has to be a better system of notation for logarithms. I think that is the main reason people freak out with them.

[@jquick85]

Still really liking this new style you're doing with the vids!

[@kristakingmath]

Thanks, Josh! :)

[@pointpleasant73]

Krista King, thank you for this good information...

[@sh0me14]

Really love the new video style, it's very refreshing. sick of the old ways. Great job.

[@liquoricecheddar]

I still don't get the concept!!!! Negative bases with odd powers will still give us a negative argument. E.g. -3^(3)= -27 . Therefore log[base -3] of -27 will be 3. This is what I think is true and why I believe that there are negative logs, but this doesn't work, not with the calculator at least(it says no real solution). Hope you answer my question and thank you. Great video BTW.... P.S. you inspired me to make math colourful.

[@rokmusic1921]

Abdullah Alzayed I found a solution for your question. Here: kzbin.info/www/bejne/qaCpdYhjZtChhKc

[@lesleemorrison1663]

This is a great video, especially for introductory lessons in high school. Makes a lot of sense. Thanks for the video :)

[@lennyadams8275]

This video is SO helpful! Thank you, Krista! :D

[@wei-kunchung9997]

Probably one of the best explanations on this topic... Thank you!

[@supercool7848]

Thanks for your excellent videos! math is the king, when i learned about systems of differential equations it opens a new chapter of my life, then i learned that there isn't just a functional relationships exists but also a statistical relationships exists

[@coffeezealot8535]

Coolest math videos ever! Love when I find these amazing math channels; they make our learning 1000 times easier and more enjoyable. Thanks for your work!

[@kristakingmath]

😊

[@sahasss7100]

Thank You so much....... I had so much confusion regarding the base and the argument, now its crystal clear. Thanks again

[@pashubhatt]

Thankyou Krista. Simply stated the complex theory and easy to grasp.

[@abdurahmanitani5982]

I love the new style of your videos, you rock Krista

[@kristakingmath]

Thanks Abdurahman, I'm so glad! :D

[@Harglin]

Honestly in my 20 years of being alive this is the most useful math video I have come across

[@patrisagar6548]

First of all that's a wonderful idea to use papers for explaining without the need of any software or something. Brilliant! Thanks

[@doukain2096]

This answer I was looking for, thanks :)

[@kamehamehaDdragon]

Great video! Very well explained! Love your videos!

[@rchiuz01]

Nice video! I like the format with with the colorful papers. Your video did bring up a thought, why do we use log base 10 the most, guess that's for Google. Thank you for the informative video 😬

[@robgordon1489]

I got it down below. Thanks and great info well explained!

[@JJ-uj1wi]

Thank you for the clear explanation!

[@jeedollee4032]

I found the most perfect tutorial video which is not decades old. Seriously bless you so much because you are saving my life!! please keep it up until I say good bye to maths ! your explanation is slow and clear especially with well prepared visuals which make me understand so well!! Thank you again!!

[@kristakingmath]

I'm not sure how long you'll be studying, but I don't plan to stop any time soon. :) I'm glad the videos are helping!

[@Mechfusion9478]

I'm doing calculus but found myself watching logs! U make math interesting Mrs Kristen

[@Pentaguin]

Animation was first introduced in 1906. People before 1906 :

[@NaeemIrum]

o yes. . 👍👍👍 excellent . . thanks for info . . good job 👏👏👏

[@efinityatkins]

Thank you you have explained this better than my teacher.

[@TheUberPeople]

i am loving these new videos!! makes me LOVE math again!!

[@kristakingmath]

Awesome, that makes me so happy!

[@jhaji8856]

mam i was confused about these properties of logarithm but after watching your video all my doubts are clear mam (love from india) i am thankful to you 🙂🙂

[@dylanmiddleton5917]

this is very helpful thank you for this!

[@sevketkeskin4715]

I am a student at Middle east tecnical university in Turkey. I could not understand that why can not be negative on argument. I have learned this. Thank you... I hope that I will be your student on the calculus class. When ı was watching this video ı said that myself I should be student in the USA....

[@Mark16v15]

According to blackpenredpen, ln(-2) = (2m+1)iπ + ln(2). Except when dealing with zero, just about any equation has a solution if complex numbers are allowed.

[@avimeryljr69HxA]

Damn, you help me a lot! Thank you for making this video❤️

[@kristakingmath]

You're welcome, Avimeryl, I'm so glad it helped! :)

[@tasninnewaz6790]

Awesome presentation and cool thinking . Krista, really a king !!!

[@kristakingmath]

Thanks, Tasnin! :)

[@djemi3etelminigeekandhacke343]

sorry but you can put a negative number on the argument of a log function (and you can even use a complex number). for exemple : log (-2) = log(2) + iπ. note that the negative number are just a complex number wish their argument is π. the log of a complex number is equals of the log of it's modulus + i times it's argument. for example : log (2i) = log(2) + iπ/2. for any complex number Z with modulus r and argument φ. Z = r times exp(i φ) log (z) = log (r times exp(i φ)) log (z) = log(r) + log (exp (i φ)) log(z) = log(r) + iφ we can even put the negative (and even complex) number in the base because log(a,b) = log(a) / log(b) when I wrote log(a,b) I mean that log of a in the base b. and log(z) in the log of z in the base e. anyway pleas post a video about the log of complex numbers.

[@saavedyv2]

Djemi3et el MINI Geek and Hackers Explaining this would take much deeper detail, but it's good to keep in mind for advanced maths

[@scitwi9164]

True, complex numbers are kind of… complex :q But it is still better to tell the students that it is POSSIBLE but they'll learn about it later, than to lie to them that this is IMPOSSIBLE :P (And I say "lie" because if you know that complex numbers exist, but deliberately hide this fact to your students, you're not making a mistake - you're lying). The former approach induces curiosity. The latter approach induces defeatism and subordination. And they may never get to the point when they'll learn about complex numbers and they will be convinced through the rest of their lives that this is impossible :P Congratulations, you made a zombie.

[@cgmiguel]

In fact she must've mentioned this is for real numbers only (to keep things simple)

[@djimonmclean164]

Thank you so much, I finally understand!!! 😊

[@kristakingmath]

Oh good! So glad it helped!

[@joaovasconcelos5360]

Bravo, thank you so much!

[@zaperizzz6659]

helped me out so much, thank youu !!!

[@kristakingmath]

You're welcome Zape! I'm so glad it helped! :)

[@CstriderNNS]

what about using imagenary numbers to solve log base (-arg)?

[@user-zq7dk4fk2y]

Thank you so much for making my math homework much easier and more understandable.

[@someonespadre]

It was common practice when logarithms were in use for computing to add 10 to negative logarithms. This is because if you look up the mantissa of a negative log it will yield the reciprocal of the number you want. So you add it to 10 algebraically, look it up, place the decimal per the characteristic then move the decimal 10 places left.

[@learningforalldsc]

It was a very good and doubt clearing vedio...thank u very much!

[@kristakingmath]

You're welcome, Deepak, so glad it helped! :D

[@abdulelahaljeffery6234]

I love your videos, they're great! at 3:00 you said "we can't take the square root of a negative number". I think you mean we can't take square root of a negative number in the real numbers ... right?!

[@kristakingmath]

Right! :)

[@MisterrLi]

As shown in the video, negative logarithms would lead to complex numbers, and so they aren't allowed by the definition (using only real numbers). Some commentators have come up with examples of negative logarithms, but those are connected to the extended logarithms where complex numbers are allowed for, see: en.wikipedia.org/wiki/Complex_logarithm

[@gfarj]

Seriously, I got shiver about how you deliver math concept..

[@kristakingmath]

:)

[@js2010ish]

Great vid 🙏🏼

[@junrancao3896]

You are awesome :D So well explained & lovely presentation

[@kristakingmath]

Thank you Junran!! :D

[@WrestlingTournamentsDotCom]

I love Logarithms!

[@benjaminbarreravasquez1990]

Excelente video, tienes gran pedagogia para enseñar, deberias traducir tus videos con subtitulos para quienes no tenemos un ingles muy fluido.....excelente trabajo y explicacion....te felicito!!!! saludos desde Chile. :)

[@kristakingmath]

Thanks! I would love to have the videos translated. I would just need help doing so. :)

[@americanswan]

Slide rules are so cool. Sadly I don't have one or have a use for one. Lots of things are awesome, but I don't need one.

[@kkamous7278]

you are the best teacher in mathematics,,,, may you live long

[@bigchrissy]

She's not the best teacher of mathematics.

[@brod515]

why are negative numbers excluded as the base for every scenario. it makes sense that log base -2 of x =1/2(any number that will cause a root of negative number), but why can't we have log base -2 of x = 3

[@carultch]

We can, just not when limited to real numbers. You can raise (-2) to integer powers, and it will alternate between positive and negatives. (-2)^1 = -2 (-2)^2 = +4 (-2)^3 = -8 (-2)^4 = +16 And when raising it to non-integer powers, there are no real solutions. You have to introduce complex numbers, to be able to have generalized logs with negative bases. It turns out, we CAN have a log with a negative base, when introducing the full picture of complex log. Given any complex number z, the natural log of it is: log(z) = ln(|z|) + i*(2*pi*k + angle(z)) where k is any integer. Note that log(z) has nothing to do with log base ten, it's all base e. A negative real number has an angle equal to pi, so given that z is a negative real number, and to keep it simple, let k=0, we'll get: log(z) = ln(|z|) + i*pi So suppose we want log base -2 of 3. This means: log(-2) = ln(2) + i*pi log(3) = ln(3) log_-2 (3) = ln(3)/(ln(2) + i*pi), and evaluating, we get log_-2 (3) = 0.07357 - 0.3335*i What if we try an actual power of 2, where we expect an integer solution? Like log_-2 (4). log(16) = ln(16) log_-2 (16) = ln(16)/(ln(2) + i*pi), and evaluating, we get log_-2 (16) = 0.1857 - 0.8416*i Ok, that's weird. What happened to the +4 we should be expecting? It turns out, we have to try other values of k, and we also have to try other values of log(16). Let ℓ be the arbitrary integer for log(16). log(16) = ln(16) + 2*pi*ℓ*i log_-2 (16) = (ln(16) + 2*pi*ℓ*i)/(ln(2) + i*pi*(2*k + 1)) By setting ℓ=2 and k=0, we can salvage the answer of +4 we were expecting. log(16) = ln(16) + 4*pi*i log(-2) = ln(2) + i*pi log_-2 (16) = (ln(16) + 4*pi*i)/(ln(2) + i*pi) = +4

[@brod515]

@@carultch Hi, thanks for replying. although this was around 7 years ago I can't even really remember what I was asking and why and I'm also feeling to lazy to do that right now. hahha. I might look at it later. but really thanks for the reply,

[@rahulwani9198]

Thank you very much ❤

[@holyshit922]

Candidate for result can be complex number but in complex domain there are some problems Exponential is periodic like trigs in real domain so inverse function can be calculated on the interval that exponential is injective Secondly in complex there is no linear order so relations such as < or > do not work

[@neelamjoshi4804]

Nice explanation

[@jagritsharma1195]

Nice explanation thx

[@pacesferry]

God bless Krista! Love these videos.

[@hg2.]

Isn't she the best math teacher on KZbin?

[@kauuu4137]

Can we allow negative bases if log_x_(z) = y if x,y belongs to integers?

[@SeeThruME4You]

The smartest human being alive.. I love this woman. she saved my behind so many times... could you please start discrete math?

[@akshaysanthosh4667]

Simply brilliant explanation

[@kristakingmath]

Thank you so much, Akshay! :)

[@subStuff]

AMAZING EXPLANATION SKILLS.

[@kristakingmath]

Aw thanks!

[@ferus5583]

Thank you so much!!!

[@kristakingmath]

You're very welcome, Ferus! :)

[@gustavopaz5453]

Krista: "Arguments of logatihms cannot be negative" Complex set: "Am I a joke to you?"

[@thecoder5939]

3:06 log -27 to the base -3 =3 then what's the problem with multiplaying it odd times like here -3 is multiplied odd times i.e.3 to give -27

[@qzorn4440]

what is a number based on, positive and negative energy? thanks.

[@hamsack981]

can you do a vid on separable equations using (x^2)+(y^2)=(k^2)

[@kennybassarath7168]

The music is also quite relaxing. I usually listen to Heavy Metal, but if I listen to metal I can't do math.

[@shadowfire04]

this saved my ass because now i get it thanks

[@thegreatdigger865]

I really like this teacher

[@lightningrain2397]

Could you make a video about Geometric Series, please? If you already have one, can you please send a link, because i cant seem to find it in your videos

[@kristakingmath]

Hi Cesar! I actually have a lot of videos about geometric series! :) kzbin.infosearch?query=geometric+series

[@lightningrain2397]

Dont know how I missed that. thanks!

[@JordanEdmundsEECS]

Heh. We can't take the square root of -2, challenge accepted. You're awesome, also if you're interested in doing videos for complex analysis that would be very cool.

[@kristakingmath]

"if you're interested in doing videos for complex analysis"... challenge accepted. :)

[@felixwinchester9256]

you explain well.subscribed.

[@kristakingmath]

Thank you!

[@surajkachary6933]

Wow lovely explaination .Tnq so much☺️

[@kristakingmath]

You're welcome, Suraj! 😊

[@rodriqcc]

I wish you had used a number greater than one for your no negative base rule. If they number was 2 instead of 1/2, would your rule still hold true?

[@syamalchattopadhyay2893]

Excellent video lecture.

[@kristakingmath]

Thanks, I appreciate that! :)

[@Chrisymcmb]

Thanks Krista!

[@kristakingmath]

You're welcome, Christopher! :)

[@charliebaker1427]

makes sense,exponents cant create a negative value as nothing times itself is negative so the argument of a log which is the product of an exponent cant be negative

[@expansivegymnast1020]

Good video. Answered my question.

[@kristakingmath]

Thanks, I'm so glad it helped! :)

[@gerardoelizondo9182]

05:00 Yes we can input 1 in the argument, always resulting in Zero: Log 1 = 0 ..... but we cannot input One as the base of the logarithm.

[@hazartullahkhan5411]

Weldon good work krista but i want to ask you a question........ Please tell about the first 10 to 20 number in logarithm table why their mean difference have two numbers given in log table while after 20 every number have only one mean difference value.please clrear it for me i am so confused about it.

[@luxithegoose4682]

Another reason why logarithm bases can't be negative is because if we think about a log function. By definition of function, x is ONLY related to one y and passes the vertical line test. And the negative bass of its inverse (exponential function, but with negative bass, it is NOT). The geometric functions inversed, DON'T PASS THE VERTICAL LINE TEST, so they are NOT considered a log function.

[@ritira20mila]

We are not discussing logarithmic functions. We are discussing logarithms.

[@roshanacademy8265]

What a method. Great ❤❤❤

[@kristakingmath]

Thank you, Roshan! :)

[@lxathu]

This title (without watching the video carefully) may be misleading. The video is correct. While the title is not right if by saying "logarithm" we mean what usually we mean, the output of the logarithm function... which definitely can be negative.

[@jacktokmaji927]

note true, what if the base is negative, and the answer isn't half (like lets say any other number than half), this means that the equation will work! so a negative base, doesn't work only with (1/2) as an answer, but it works with other numbers, am I right?

[@karthik_silkroads]

never figured you for hip hop :)

[@sheikhbabar8927]

But if we have base -2 and argument -8 answer is 3 which is correct

[@robgordon1489]

Awesome music! What is it?

[@sparkyu2166]

So why did i do log base 10 (3) equals to -0.5 ??? The base is positive but the answer is negative

[@davidwalker8658]

Great vid thanks

[@kristakingmath]

Thanks!

[@Healthy-365]

In my school math books I saw log (0) base 10 is 1....can you please explain it ma'am ..

[@419er]

can someone explain why you cant do log base i. eg i^2 = -1 so logi(-1) should give 2. Or log -1 (i) should give 1/2 and log -2(4) should give 2

[@carultch]

You can. There is a generalized complex logarithm, that works for any complex number other than zero. It is just more complicated, which is why it is beyond the scope of a high school level introduction to logs. And this works for both the base and the argument of the log.

[@debottomroy3334]

Well I am in standard 9 and we have all those problems with log where you are to find X and used to find the square root of something and square root ends with a positive and negative value in all answer I cannot see the negative values I was wondering why it wasn't accept negative value now I realise why because it is ok with complex number that is undefined and I am in standard 9 I don't need that complex number so this helped me a lot

[@gautamsah5075]

At 2:52, If you had taken log-2(X)=2 (instead of 1/2) ,then wouldn't you get (-2)^2=4 (=X) . I meant, in Log-2(4) =2 ,where base is negative, eventhough answer is defined as (-2)^2=4. Then why it is said that base cannot be negative.

[@devondevon4366]

She explains it well.

[@yaeg]

but what if the base number is negative and the solution is a positive integer?? wouldn't that be possible? and if so, why don't we use negative base numbers?

[@myeshafarzanatahi9154]

Thank you so much

[@kristakingmath]

You're welcome, Myesha! :)

[@thespiciestmeatball]

I have a question. Is it valid to say that since e^(iπ)=-1 and e^ln(x)=x then ln(-1)=iπ?

[@carultch]

It is valid. In general, complex log of any complex number other than zero, is given by: log(z) = ln(|z|) + i*(angle(z) + 2*pi*k) Where k is any arbitrary integer. To do a log of a complex number z with a base (b) that could be any complex number other than zero, the formula is: log_b (z) = [ln(|z|) + i*(angle(z) + 2*pi*k)] / [ln(b) + i*(angle(b) + 2*pi*ℓ)] Where k and ℓ are two arbitrary integers that are independent from each other. When b is a negative real number, this reduces to: log_b (z) = [ln(|z|) + i*(angle(z) + 2*pi*k)] / [ln(|b|) + i*pi*(2*ℓ + 1)]

[@tranhoangkimlong4624]

Excuse me I have a question at 3:38, Why can't base be negative? log1(1)=1, log0(0)=0 or log -2(-8)=3 still make sense so that why the base of log is define as the number except 0 ,1 and negative. And log5(1/5)= -1( negative)

[@carultch]

Logs of negative numbers do exist, you just need to enter the world of complex numbers to make sense of them. To keep it simple for a high school level, we just "lie" and say they don't exist.

[@300PIVOTMASTER]

log[base -3] of -27 is undefined by calculators, but -3^3 is definitely -27. What do

[@kuasocto3528]

I dunno, but it fucks me over as well

[@hanlly5559]

by definition base > 0, base != 1 and x > 0

[@carultch]

In general, for any complex number z: log(z) = ln(|z|) + i*(angle(z) + 2*pi*k) For -3 & -27, where k & ℓ are independent integers: log(-27) = ln(27) + i*pi*(2*k + 1) log(-3) = ln(3) + i*pi*(2*ℓ + 1) Simplify: log(-27) = 3*ln(3) + i*pi*(2*k + 1) Construct the ratio, to find log_-3 (-27): log_-3 (-27) = log(-27)/(-3) = [3*ln(3) + i*pi*(2*k + 1)]/[ ln(3) + i*pi*(2*ℓ + 1)] Modify the numerator, so it is proportional to the denominator, and can cancel to equal a real number at the end of the day. Comparing real parts, we see that the real part of the top is 3 times the real part of the bottom. So this means, the imaginary part must also be 3 times the imaginary part. Due to the fact that k & ℓ are arbitrary, we can set it up so this can happen. (2*k + 1) = 3*(2*ℓ + 1) Solving for k, we get the general relationship between them, k = 3*ℓ+1. To keep it simple, let ℓ=0, implying that k=1. Now, we can do the following: [3*ln(3) + i*pi*(2*1 + 1)]/[ ln(3) + i*pi*(2*0 + 1)] [3*ln(3) + 3*pi*i]/[ln(3) + i*pi] Factor out the 3: 3*[ln(3) + i*pi]/[ln(3) + i*pi] = 3 Which is precisely what we were expecting, since (-3)^3 = -27.

[@ramkrishnajoshi9297]

i have a very basic question. What if we say log of -8 to the base -2? In this case we can find the answer to be 3 because -2^3 = -8. Also after some research I found that the natural log of -1 is defined and is pie*i. So how can u say that the argument cannot be negative?

[@cgmiguel]

Because It only works for certain numbers

[@sci-fithinker875]

since the base a

[@carultch]

You can take the logs of negative numbers, it just becomes more complicated, because you have to introduce complex numbers to make it work. Consider your example of log_-2 (-8). To unpack this, we need log(-8)/log(-2), where log is complex log base e. In general, the log of any complex number z is: log(z) = ln(|z|) + i*(angle(z) + 2*pi*k) where k is any arbitrary integer. Given a negative base b, this means: log_b (z) = [ln(|z|) + i*(angle(z) + 2*pi*k)]/[ln(|b|) + i*pi*(2*ℓ + 1)] where k & ℓ are independent arbitrary integers. Let's try it for z=-8 and b = -2: log_-2 (-8) = [ln(8) + i*pi*(2*k + 1)]/[ln(2) + i*pi*(2*ℓ + 1)] Keeping it simple and letting k and ℓ both be zero: log_-2 (-8) = [ln(8) + i*pi]/[ln(2) + i*pi] log_-2 (8) = 1.093 - 0.4208*i Huh? A complex result, how did that happen? Shouldn't it be 3? Turns out, it will equal 3, if we set k=0 & ℓ=1. Just as we were hoping. But we have to find the right combinations of the arbitrary integers to get this to happen. It also happens at k=1 and ℓ=1

[@Hala-rv2ic]

cant we just do log base 0 of 0 then it is defined and equal to one i really dont get it