Oon Han unless 1/log(a)+1/log(b) = 1/log(c) In which case x is any real number
@ffggddss5 жыл бұрын
@@NoNameAtAll2 Yes! Except that in that case, x is any *positive* real number. Fred
@UltraLuigi24015 жыл бұрын
@@ffggddss Well really any nonzero real number, since negative logs exist, they're just not real.
@ZmileFeat6 жыл бұрын
0+0=0 correct
@xin49415 жыл бұрын
This is the real result.
@brenowemanoel23545 жыл бұрын
0+0 = 1
@brenowemanoel23545 жыл бұрын
Something
@U014B5 жыл бұрын
But did you also know that 0 = 0 + 0?
@CatchyCauchy4 жыл бұрын
@@U014B can you prove it?
@mathewheffley1216 жыл бұрын
Why does he have a Death Star replica in his hand?
@lh16905 жыл бұрын
Class control.
@dionysianapollomarx5 жыл бұрын
He engineered it.
@maximilianludwig43015 жыл бұрын
Your maths teacher skills has shown me the beauty of maths. Thank you master
@andresortega20986 жыл бұрын
1/log(5) + 1/log(7) - 1/log(25) != 0, proof by intimidation
@gaymerjerry5 жыл бұрын
if you want a further proof lets assume it is equal to 0 1/log(5) + 1/log(7) - 1/log(25) = 0 1/log(5) + 1/log(7) = 1/log(25) (log(5)+log(7))/(log(5)log(7)) = 1/log(25) log(5)log(7) = (log(5)+log(7)) log(25) log(5)log(7) = (log(5)+log(7)) log(5^2) log(5)log(7) = 2 (log(5)+log(7)) log(5) log(7) = (log(5)+log(7)) 2 log(7) = 2 log(5)+ 2 log(7) 0 = 2 log(5)+ log(7) log(5)>log(1) therefore log(5)>0 log(7)>log(1) therefore log(7)>0 therefore 2 log(5)+ log(7)>0 therefore 1/log(5) + 1/log(7) - 1/log(25) is not 0
@polyhistorphilomath5 жыл бұрын
Multiply through by 2 log(5) to clear the fraction on the right first. Then 2 (1+log(5)/log(7)) log(x) = log(x) obtains. Subtract log(x) from both sides. [1+2 log(5)/log(7)] log(x)=0. To clarify, convert from the form log(a)/log(b) to the equivalent log_b(a). [1 + 2 log_7(5)] log(x) = 0. Let y = 1 and let z = 2 log_7(5). y is a positive quantity. Rewrite z=2 log_7(5) as log_7(25). log_7(7^1) < z < log_7(7^2) because 7^1=7 < 25 < 7^2=49. Therefore 1 < z < 2. The expression in question is the coefficient of log(x). It is equivalent to y + z. Add 1 to the inequality. 1+1< y+z < 2+1. 2 < y+z < 3. 0 < 2. 0 != y + z. QED
@mwitamgesi65965 жыл бұрын
Andrés Ortega 39
@holhenrik5 жыл бұрын
Enter it into a calculator, you'll get 1.8... anything. 1.8 != 0 so the expression != 0.
@fstasel6 жыл бұрын
Here is the problem: Is there a non empty set of integer triplets (a, b, c) such that 1/log(a)+1/log(b)-1/log(c)=0?
@thebackyardmovies2 жыл бұрын
The set (c^2, c^2, c) for all positive integers c works
@espadadearthur1174 Жыл бұрын
@@thebackyardmovieswhat if we don't allow duplicates, that is: a =/ b =/ c
@qwertyuioph Жыл бұрын
@@espadadearthur1174 then its just all c^x c^y & c^z where 1/x+1/y=1/z eg 1/2+1/2=1/1 1/3+1/6=1/2 etc
@MrJdcirbo5 жыл бұрын
I spent an hour in this and I came to the same conclusion. I thought I was doing something wrong lol
@deeptochatterjee5326 жыл бұрын
What a troll
@aidenwinter11175 жыл бұрын
This is probably the biggest math troll after Fermat’s Last Theorem
@MathZoneKH5 жыл бұрын
Mr prof. Can ask u s.th? b we call is a base of log and what is x what do we call ????
@shaunderoza23217 жыл бұрын
It would be more interesting if you had a constant term on the right hand side so the solution is non-trivial.
@blackpenredpen7 жыл бұрын
I know. But that was the question that the viewer asked.
@jaimemartins84805 жыл бұрын
Great vídeo, great equacion, great example. Thanks for this vídeo
@blackpenredpen5 жыл бұрын
Thanks!
@chanchalsingh86643 жыл бұрын
Very good bro ,you really explained very well
@Cannongabang7 жыл бұрын
x=1 hahahah
@dalek10997 жыл бұрын
well it has important significance because log(1)=0 in any base and thats important as the equation he wrote could be written in any base so it really had to be x=1 as the solution.
@polyd25007 жыл бұрын
Thanks for your helpful videos!
@shay_playz9 ай бұрын
I used a completely different method. I used reduction to smaller possible base so log4X is 1/2Log2X hence we multiply x½ with x¹ to get x^(3/2) then we get 3/2log2X =6 therefore log2X is (6×2)/3 which is 4 then eliminate log so x is 2⁴ which is 16.
@CalculusPhysics5 жыл бұрын
i guess this solution shouldn’t be surprising. if you think about the graphs for these functions, the only place they will cross is at x=1
@NonTwinBrothers3 жыл бұрын
That ending really got me lmao
@alyciamg4 жыл бұрын
okay logx becomes 1 but what about to the right, everything you're adding in the other bracket?
@saidhelal58666 жыл бұрын
Another one with lovely answer find value of x if, log_5 (x)+ log(x/2)=log_√(5)(x)
@boium.7 жыл бұрын
I have a kind of similar question now. Is it possible to write log5(a)+log7(a) as log25(b) for any positive number a?
@tibimose8237 жыл бұрын
no, only for a=1. While watching the video, I was tempted to cancel out log(x), which actually means dividing by 0 after seeing the solution. If you do that, you get some logs which are not equal to each other (as he pointed out in the video that the second term cannot be zero)
@leonardovasquessailer16537 жыл бұрын
His question is a bit different. You can indeed write log5(a)+log7(a) as log25(b) for any positive a
@leonardovasquessailer16537 жыл бұрын
Tibi Mose All you are going to need is to have b = 25^[log5(a)+log7(a)]
@boium.7 жыл бұрын
Leonardo Vásques Sailer sorry but I've phrased my question poorly. I ment 'can you find b algebraicly?' What you said helped me though because b=25^(log5(a)+log7(a)) can't be solved algebraicly. (I think)
@Kishore82195 жыл бұрын
So for any different base x is 1 is it right
@thomasg68307 жыл бұрын
I think its somewhat odd that you prefer log10(x) and power10(x) over ln(x) and exp(x). Thats like using tau instead of pi.
@otherodd2 жыл бұрын
I think in most cases, e is overused. In this example, you deal with integers as bases and answers so using e would (potentially) just yield irrational results you cannot do anything with. (In many cases natural log and exp is very useful tho)
@dhyeypatel83997 жыл бұрын
Pls make a video on limits and derivatives
@blackpenredpen7 жыл бұрын
I have old vids, Please go to blackpenredpen.com and calculus ressources
@toddbiesel42887 жыл бұрын
So the three bases don't have to be 5, 7, & 25. They can be any three positive numbers other than 1, like pi, e, & i^i.
@deeptochatterjee5326 жыл бұрын
Todd Biesel Well you have to be careful. If you do 25,25,5 then you get all numbers that the function is defined for
@spinalcord45292 жыл бұрын
does it make sense to factor 0 out of an equation?
@gamingwithlegends7006 Жыл бұрын
Yesn't
@BoyetPagalan-lq3ri11 ай бұрын
How to find the x in this , log base 2 of 3(2x - 1) = log base 3 of 3x - 2?
@mukaddastaj52233 жыл бұрын
Thank u!!! But how ro solve this? (x-1)Log3_(x)=x+1)/2
@tokajileo59285 жыл бұрын
in Europe we use for log base 10 the sign lg ( lg(10)=1) ln is natural log and we use log only is base if not e or 10.
@agfd56595 жыл бұрын
I also live in Europe, but we use red pen's notation here
@Kitulous5 жыл бұрын
@@agfd5659 in Russia we use lg, tg, ctg, etc. Instead of log, tan, cot.
@agfd56595 жыл бұрын
@@Kitulous here, in the Czech republic, we use log for decadic logarithm, ln for natural logarithm, tg for tangens, cotg for cotangens.
@devamrh2 жыл бұрын
Hey dude. Stuck with this problem can you help me out? What are the roots of the function f(x) = (log(3x ) − 2 log(3)) · (x 2 − 1) with x ∈ R?
@farhanfouadacca7 ай бұрын
Amazing thing is the handling of micish thing in his hand.
@jackchung7 жыл бұрын
Would you mind adjusting the camera's angle please? There are several light spots on the whiteboard that cause your writings difficult to read.
@faith31747 жыл бұрын
Could you please evaluate ∫ √[1+2cot(x)·(csc(x)+cot(x))] dx
@Haalita213 жыл бұрын
The trick is to substitute in the Pythagorean identity of 1 = csc²(x) - cot²(x). Upon expanding the brackets and utilising the substitution, the inside expression transforms to csc²(x) + 2csc(x)cot(x) + cot²(x) which can be further simplified as [csc(x) + cot(x)]². The square cancels with the square root, and you’re simply left to integrate csc(x) + cot(x).
@Euphist157 жыл бұрын
Trivial
@dcs_07 жыл бұрын
yeah, youre right, a bit boring. I would have loved it if there was some elegant, non-trivial solution tbh
@blackpenredpen7 жыл бұрын
i know. But that was the question that the viewer asked.
@ViralVision7 жыл бұрын
How do you do the derivative of x^x?
@lumpysparrow3397 жыл бұрын
Rewrite it as e^ln(x^x). You can break this apart as e^(xlnx). Differentiating using the chain rule gives e^(xlnx)*(lnx + 1). Since e^(xlnx) is equal to x^x, you can get write the answer as: x^x(lnx + 1), and that's the answer edit: I should mention that you initially rewrite it using e because it gives an easy way to break apart the two x's
@blackpenredpen7 жыл бұрын
It's here kzbin.info/www/bejne/ol7MfZpmbN-UrMU
@MrKnivan6 жыл бұрын
Implicit differentiation
@alephnull40445 жыл бұрын
@@lumpysparrow339 What you said in the edit is actually quite misleading and a common misconception in this problem. If we forget about e for a second and just go back to the original question, what does it even mean? More specifically, what the heck does something like pi^pi mean? If you can't even give a definition of x^x (for all real numbers x), then you definitely can't take its derivative. In fact e^xlogx is the very definition of x^x, and that is why writing it like that magically makes it easy to differentiate.
@KrzaqDBP7 жыл бұрын
log_x^n(p)=1/n log_x(p) :p 25=5^2
@jennagrace71055 жыл бұрын
What if I have just a normal number on the other side of the equation? The question I'm stuck on is log4X + log2X = 6 if anyone could help that would be great
@tyromejenkins24425 жыл бұрын
Your prayers have been answered 🙏🏽
@XZellTheBest5 жыл бұрын
i suppose you mean "log base 4 of X plus log base 2 of X is equal to 6", right? Or you mean "log base 10 of 4X plus log base 10 of 2X is equal to 6"? You can write "log base 4 of X" as "log base 2 of X over log base 2 of 4" right? This is just "log base 2 of X, all over 2". So the sum is (3/2)*log base 2 of X. This is equal to 6. That means that the log base 2 of X is equal to 6*2/3=4. So X=2
@daffadaffa80765 жыл бұрын
@@XZellTheBest isn't x=16?
@_Dalember_5 жыл бұрын
When you add logarithms with the same base, is the same as multiplying them , so this would be: log(4x*2x)=6 , so: (10^6) = 8(x^2 ), so: x= √(10^6/8)
@muhammadazeemkhan17624 жыл бұрын
Is there a particular solution to this problem besides 1
@johnmanuelm.escueta69875 жыл бұрын
Can you show why tthe sum of thos is log base 25 x
@darcorbit127 жыл бұрын
Can you please tell me why we have problem with 0^0, if lim(x^x)=1, when x is going to 0?
@stephenmontes3497 жыл бұрын
Denis Dolich you use logarithm properties to solve it
@ugurcansayan7 жыл бұрын
Think it like this: 0^a = 0 b^0 = 1 so, 0^0 = ?
@Hauketal7 жыл бұрын
Denis Dolich lim(x^0)=1, but lim(0^x)=0 for x→0. None of them is better, so 0^0 is left undefined.
@darcorbit127 жыл бұрын
exactly, does it make any problem?
@darcorbit127 жыл бұрын
Ok, i don't know how to get that answer (lim(0^x)=0 for x→0). But i cheked it in program, and it's true. I understood your version, but answer for lim(x^x) is still 1 like i solved, and like the same program says. Even calculating using very small numbers you will get approximately 1. Is it problem of using just very approximately values, but not actualy zero...
@m.raedallulu41665 жыл бұрын
Brilliant ! Thank you so much!
@thefootboy5537 Жыл бұрын
Thank you so much
@swarm41967 жыл бұрын
Can you explain us how to calculate log10(x) without a calculator or it's definitely impossible?
@pfeffer17297 жыл бұрын
You can get arbitrarily good precision with a Taylor Series.
@somebodysomewhere92536 жыл бұрын
It can also be done by iterative process (how calculators work it out in the first place) Set a step variable to 1 Set a power value to one Find 10^power Is 10^power greater than x? If so, times step by -0.5 Now change power to be power + step. Go back to the step telling you to find 10^power When you reach your degree of accuracy, output the power variable.
@TalestoryJL6 жыл бұрын
Hi, I have a question! at 3:40 you've explained how to solve the log base 10 of x = 0 by adding 10 as a base of their power on both sides. Could you explain a little further? about how this solution is cancelling out the log base 10. I mean, could anyone give me a tip Why the action of giving 10 as a base will cancel out log base 10. I get the process, and I understand what to do when similar question comes up in the future, but I'm questioning the understanding part. I want to know "why" it's done that way.
@MrKnivan6 жыл бұрын
Talestory JL idk if this answers your question but basically anytime we have log(x)=y, we’re saying “what number can we raise 10 to in order to yield y?” So if we take 10^[log(x)] we are effectively taking 10 and raising it to the number that yields us y. So the log(x) and 10 are “canceled out” in a sense because all we did was change the whole expression of 10^[log(x)] to y. Does that make sense?
@Justiin_rm4 жыл бұрын
How about complex solutions?
@jaimemartins84805 жыл бұрын
Ótimo vídeo, ótima equação, ótimo exercício. Muito interessante
@saidhelal58666 жыл бұрын
change to think for small challenge try to solve: 10^log(x+5)+100^log(x+3)=100000000^log√(2)*√(10)^log(49)
@armenstaubach92763 жыл бұрын
Thanks, that’s awesome...
@Nickesponja5 жыл бұрын
What if you had log_5(z)+log_7(z)=log_25(z) where log_x(z) denotes the branch of logarithm in which the argument of z is taken to be between x and χ+2π?
@kimlatifa12293 жыл бұрын
Can you solve this inequaliti please log₃(x+2)
@moudar1237 жыл бұрын
Thank you very much, your video was very helpful for me even if i am a college student!!
@danny1504-g2d9 ай бұрын
I really apprieciate your work. Hope you can help me with this fun math problem, it's from my math teacher: Given x+y=3 and x^x + y^y = x^y + y^x Find x and y. Thank you!
@sandeepshastry66477 жыл бұрын
Can you show me the derivation of ln(x)=2.303log(x) pls
@lumpysparrow3397 жыл бұрын
d/dx[lnx] = d/dx[2.303logx] 1/x = 2.303/x (log and ln have the same derivative as long as log is base 10. If it is not base 10, first rewrite using change of base)
@abdulazizalbaiz27587 жыл бұрын
ln(x)=log(x)/log(e) 1/log(e)=2.30258...... So, ln(x)~= 2.303log(x)
@sandeepshastry66477 жыл бұрын
Sasuga pls read my question properly. I did not ask for derivative, I asked derivation.
@deeptochatterjee5326 жыл бұрын
Sasuga You people never learn (/s). The verb for taking a derivative is DIFFERENTIATE not DERIVATE. DERIVATION REFERS TO DERIVING SOMETHING FROM PREMISES.
@alxjones6 жыл бұрын
@@deeptochatterjee532 Not so fast. A derivation is also a differential algebraic operator D such that D(ab) = D(a)b + aD(b), and derivative is such a derivation. However, it is almost always possible to tell from context when someone is referring to an abstract derivation.
@pituitlechat38077 жыл бұрын
Hi all Here we take a=5, b=7 and c=25. It's seem strange to me that if you use others values for a, b or c you can have a second solution? Perhaps, in all case there are more than one solution in the complex number's world?
@renzalightning60087 жыл бұрын
Are you referring to the bases of the logs? if so, I think it unlikely for x to be anything other than 1 in a system like this. I might be wrong but it seems that way XD
@manudude027 жыл бұрын
well, if a=125, b=25 and c=5, then x=125 works. When c is the highest value however, there can never be a positive non-zero answer edit: b should be 5^(3/2) as geCeeMeS says below, though the logs don't become 0s, you are effectively trying to solve x+2x=3x though.
@GeCeeMeS7 жыл бұрын
Well, not quite. If you choose e.g. a=125, b=5^(3/2) and c=5, then the term full of logs becomes zero, and the equation is true independent of the value of x.
@pituitlechat38077 жыл бұрын
in fact I found that if c = e^(ln(a)ln(b)/ln(ab)) then log_a(x)+log_b(x)=log_c(x) for all x. And effectively c can be > a or b if a or b is
@Mahi-07-075 жыл бұрын
Thanks sir love from india
@theophonchana50253 жыл бұрын
log base 5 (x) = log (x) ÷ log (5)
@chelsea28074 жыл бұрын
Thanks bro!!!
@gesucristo07 жыл бұрын
Could you use feynman's trick to solve integral of 1/ln(X) please?
@srijanraghunath46425 жыл бұрын
Easier said than done If integrating ln(x) from 2 to t could be done, we would know much more about the distribution and number of primes within a given boundary
@mikemangilashi4487 Жыл бұрын
Thank you❤
@theophonchana50253 жыл бұрын
log base 5 (x) + log base 7 (x) = log base 25 (x) log base 5 (x) + log base 7 (x) - log base 25 (x) = 0
@quasar_catfish7 жыл бұрын
He forgot the closing parenthesis on the log(25) in the step before last... It annoys me =P
@NoNameAtAll27 жыл бұрын
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@Gawkie7 жыл бұрын
i got x= 7/5 when i converted the logs to lns, heres how i did lnx/ln7+lnx/ln5=lnx/ln25 e^(lnx/ln7+lnx/ln5)=e^(lnx/ln25) e^(lnx/ln7).e^(lnx/ln5)=x/25 x/7.x/5=x/25 x^2/35-x/25=0 x(x/35-1/25)=0 either x=0 or x/35-1/25=0 x/35=1/25 x=7/5 any idea where i went wrong??
@rigoluna14917 жыл бұрын
log_b(X)/log_b(P)+log_b(X)/log_b(Q)=log_b(X)/log_b(R) b^(log_b(X)/log_b(P)+log_b(X)/log_b(Q))=b^(log_b(X)/log_b(R)) X/P . X/Q=X/R X^2/PQ=X/R X^2/PQ-X/R=0 X(X/PQ-1/R)=0 X=0 or X=PQ/R Interesting, I haven't a clue where you went wrong (assuming you did) but it seems to work with any base
@abdulazizalbaiz27587 жыл бұрын
Your mistake is: e^lnx/ln25 does not equal x/25 and I see your confusion, In general: ln(a/b) = ln a - ln b But in your case it is (ln a/ln b) Which can be expressed as ln_b (a) But not a/b Hope that helps
@Gawkie7 жыл бұрын
Abdulaziz Albaiz yeah i realized it after posting xD great to see a fellow Arab on this channel bte
@Inujasa886 жыл бұрын
From the first look it's obviously x=1. It's so trivial!
@rubenramos38587 жыл бұрын
Can you help me with an interesting calculus problem? Its the integral of (x*lnx)^50 from 0 to 1.
@chasr18437 жыл бұрын
I tried substituting u = ln(x), x = e^u and du = dx/x ==> dx = e^u du then x*ln(x) becomes u*e^u and you have to integrate (u*e^u)^50 * e^u du = (u^50 * e^51u) du. You could try integrating by parts, but don't. Type the equation into wolfram alpha and notice their answer is some mess of a series that goes on until computation time gets exceeded. I suspect it's one of those integration-by-parts that keeps going around in circles, probably either 50 or 51 or infinity times. Are you sure you stated the problem correctly?
@OnTheThirdDay7 жыл бұрын
Note that the wolfram alpha ugly expression: www.wolframalpha.com/input/?i=int_0%5E1++(x*lnx)%5E(50)+dx is the indefinite integral in general. I would (on the basis of experience) distribute the power of 50 to both terms and try integrating by parts... try to find a pattern. Think about why the integral form 0 to 1 makes sense but in general it is really ugly.
@chasr18437 жыл бұрын
That's what I did David Herrera. I also put the definite integral into alpha. It's just a much of a mess. I also tried the integral with a power of 1 (easy) and 2 (annoying) and n. Doing it for n=50 looks hopeless to me unless you have been given a life sentence of solitary confinement in jail. The reason I transformed it before distributing the powers is I'm not sure what to do with (ln(x))^50 ( =/= ln(x^50) btw).
@davidross34877 жыл бұрын
Is log base i a legitimate function? If so is it useful?
@chasr18437 жыл бұрын
Yes it's legit. If L = log_i(x) that means i^L = x. Then substitute i = exp(i*pi/2) and you wind up with x = exp(i*L*pi/2)
@coolparrot94326 жыл бұрын
nice video!
@gamabuga6 жыл бұрын
but what do we do when constant part is equal to zero?
@primthon95966 жыл бұрын
if the constant part is zero the whole equation is true, no matter what your x is. But note that x can only be positive, so x would be in (0; ∞).
@weebql6766 жыл бұрын
The restriction of logs is that the base are different of 1, or i am wrong?
@earthbjornnahkaimurrao95426 жыл бұрын
are there other complex solutions?
@TheSandkastenverbot2 жыл бұрын
No, 1 is the only solution. The natural logarithm of a complex number z=a*exp(i*phi) is ln(z) = ln(a) + i*phi. Since we're looking for ln(z)=0 the only solution is ln(a) = 0 and phi=0, so a=1 and therefore z=1*exp(i*0)=1
@naimulhaq96264 жыл бұрын
I like blackpenredpen.
@johnmanuelm.escueta69875 жыл бұрын
Thanks you
@memememe24886 жыл бұрын
thanks
@nk-qy2xp4 жыл бұрын
x=1?!?! I've been bamboozled!
@jotakiepie5 жыл бұрын
thank you omg
@saimachy83605 жыл бұрын
I didn't get the note... how did that come!!
@seenapushparaj62436 жыл бұрын
Tnx!
@theophonchana50253 жыл бұрын
#logarithm #log #logbase5 #logbase7 #logbase25
@pushkarajsalunke34616 жыл бұрын
Just take the base as X and you will understand the problem and why it should be 1 intuitively.
@goodboypi56155 жыл бұрын
I hope can you help me in this E.Q. Log_3(x)+log_9(x)=6 thanks Soumyadip.
@spamspam94955 жыл бұрын
Log(X)=4 is the answer
@buongiorno52334 жыл бұрын
@@spamspam9495 with a base of 3, right?
@dhay39826 жыл бұрын
But Log_5(x+1)+log_7(x-1)=log_25(x)?
@rezamiau5 жыл бұрын
well done! thank you.
@1966lavc3 жыл бұрын
without solving the equation , you can tell x=1 is the solution
@no_de2 жыл бұрын
I think just making all of the logarith into log5 is way too ezier.
@jeffreyluciana87114 жыл бұрын
Wonderful
@eduardoyotana5196 жыл бұрын
if we have different bases and different logarithms. how i can do? Like x, y, z.... help me please black pen red pen
@pfeffer17297 жыл бұрын
You can show that 1/log sum is not 0 by using log 25 = 2 log 5, so that 1/log 5 - 1/log 25 = 1/log 25. Then you have a sum of two positive numbers.
@toripuru00697 жыл бұрын
make log with base i
@habouzhaboux94885 жыл бұрын
Log_i(e) is 2/πi is a neat formula if you want. You can conclude it by saying log_i(e) = ln(e)/ln(i) = 1/ln(i) = 1/(πi/2) = 2/πi
@habouzhaboux94885 жыл бұрын
@@polyhistorphilomath you can check the answer in wolfram alpja
@JoseFernandes-js7ep4 жыл бұрын
A mathematician using base-10 logarithms?!
@pandas8965 жыл бұрын
Only at x=1
@mahmoudashraf66015 жыл бұрын
I need the answer of this question please the integration of x^x
well because: 1/log5 + 1/log7 -1/log25 = [log(7)log(25) + log(5)log(25) - log(5)log(7)] / [log(25)log(7)log(5)] So the numerator [log(7)log(25) + log(5)log(25) - log(5)log(7)] musst be 0. But log(7)log(25) + log(5)(log(25)-log(7)) > 0 because log(7)log(25) > 0 and log(25)-log(7) > 0 and a positive plus a positive is still positive which can not be zero.
@deeptochatterjee5326 жыл бұрын
niroo s You could just put it in a calculator
@sulapanideb31182 жыл бұрын
The question is wrong because log to the base 1 is undefined.
@dhruvagoyal99455 жыл бұрын
What's with the easy questions man?
@ericdaniello56594 жыл бұрын
You can see that, whatever the solution is, raising that number to any power is also a solution because of log rules and division Any other solution than the trivial one would mean all numbers are a solution, so I narrowed down to one or all (literally) before doing the math🍻
@blazify64125 жыл бұрын
Ah yes, the loyarithm.
@bayc42077 жыл бұрын
15th
@kritikkaushal63055 жыл бұрын
lifesaver
@simpletn6 жыл бұрын
What the actual hell...
@brunoandrades55307 жыл бұрын
Could u pls integrate |x|/(|x|x - x) I do really like your videos!