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
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.
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)
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)
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
@@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.
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.
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
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
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)
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...
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.
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?
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.
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).
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?
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
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)
Sasuga You people never learn (/s). The verb for taking a derivative is DIFFERENTIATE not DERIVATE. DERIVATION REFERS TO DERIVING SOMETHING FROM PREMISES.
@@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.
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
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?
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
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.
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.
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??
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
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
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!
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?
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.
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).
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🍻
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.
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log_a(x)+log_b(x)=log_c(x)
x=1
ALWAYS
yes
Oon Han unless 1/log(a)+1/log(b) = 1/log(c)
In which case x is any real number
@@NoNameAtAll2 Yes! Except that in that case, x is any *positive* real number.
Fred
@@ffggddss Well really any nonzero real number, since negative logs exist, they're just not real.
0+0=0 correct
This is the real result.
0+0 = 1
Something
But did you also know that 0 = 0 + 0?
@@U014B can you prove it?
Why does he have a Death Star replica in his hand?
Class control.
He engineered it.
What a troll
Your maths teacher skills has shown me the beauty of maths. Thank you master
I spent an hour in this and I came to the same conclusion. I thought I was doing something wrong lol
This is probably the biggest math troll after Fermat’s Last Theorem
1/log(5) + 1/log(7) - 1/log(25) != 0, proof by intimidation
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
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
Andrés Ortega 39
Enter it into a calculator, you'll get 1.8... anything. 1.8 != 0 so the expression != 0.
@andresorega That was funny! You win the internet today.
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?
The set (c^2, c^2, c) for all positive integers c works
@@thebackyardmovieswhat if we don't allow duplicates, that is:
a =/ b =/ c
@@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
x=1 hahahah
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.
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
It would be more interesting if you had a constant term on the right hand side so the solution is non-trivial.
I know. But that was the question that the viewer asked.
That ending really got me lmao
Amazing thing is the handling of micish thing in his hand.
Great vídeo, great equacion, great example. Thanks for this vídeo
Thanks!
Thanks for your helpful videos!
Mr prof. Can ask u s.th? b we call is a base of log and what is x what do we call ????
Very good bro ,you really explained very well
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?
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)
His question is a bit different. You can indeed write log5(a)+log7(a) as log25(b) for any positive a
Tibi Mose All you are going to need is to have b = 25^[log5(a)+log7(a)]
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)
okay logx becomes 1 but what about to the right, everything you're adding in the other bracket?
Trivial
yeah, youre right, a bit boring. I would have loved it if there was some elegant, non-trivial solution tbh
i know. But that was the question that the viewer asked.
How do you do the derivative of x^x?
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
It's here ruclips.net/video/l-iLg07zavc/видео.html
Implicit differentiation
@@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.
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.
Another one with lovely answer find value of x if,
log_5 (x)+ log(x/2)=log_√(5)(x)
Pls make a video on limits and derivatives
I have old vids,
Please go to blackpenredpen.com and calculus ressources
does it make sense to factor 0 out of an equation?
Yesn't
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
Your prayers have been answered 🙏🏽
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
@@XZellTheBest isn't x=16?
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)
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.
I also live in Europe, but we use red pen's notation here
@@agfd5659 in Russia we use lg, tg, ctg, etc. Instead of log, tan, cot.
@@Kitulous here, in the Czech republic, we use log for decadic logarithm, ln for natural logarithm, tg for tangens, cotg for cotangens.
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.
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)
So for any different base x is 1 is it right
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.
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
Brilliant !
Thank you so much!
Ótimo vídeo, ótima equação, ótimo exercício. Muito interessante
Thank you❤
Thanks sir love from india
Can you please tell me why we have problem with 0^0, if lim(x^x)=1, when x is going to 0?
Denis Dolich you use logarithm properties to solve it
Think it like this:
0^a = 0
b^0 = 1
so,
0^0 = ?
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.
exactly, does it make any problem?
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...
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.
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?
Can you explain us how to calculate log10(x) without a calculator or it's definitely impossible?
You can get arbitrarily good precision with a Taylor Series.
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.
Thank you so much
Would you mind adjusting the camera's angle please? There are several light spots on the whiteboard that cause your writings difficult to read.
Could you please evaluate
∫ √[1+2cot(x)·(csc(x)+cot(x))] dx
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).
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?
Thanks, that’s awesome...
Thanks bro!!!
How to find the x in this , log base 2 of 3(2x - 1) = log base 3 of 3x - 2?
He forgot the closing parenthesis on the log(25) in the step before last... It annoys me =P
)
log_x^n(p)=1/n log_x(p)
:p 25=5^2
Could you use feynman's trick to solve integral of 1/ln(X) please?
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
How about complex solutions?
Is there a particular solution to this problem besides 1
Can you show why tthe sum of thos is log base 25 x
x=1?!?! I've been bamboozled!
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
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?
change to think for small challenge try to solve:
10^log(x+5)+100^log(x+3)=100000000^log√(2)*√(10)^log(49)
Can you show me the derivation of ln(x)=2.303log(x) pls
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)
ln(x)=log(x)/log(e)
1/log(e)=2.30258......
So, ln(x)~= 2.303log(x)
Sasuga pls read my question properly. I did not ask for derivative, I asked derivation.
Sasuga You people never learn (/s).
The verb for taking a derivative is DIFFERENTIATE not DERIVATE. DERIVATION REFERS TO DERIVING SOMETHING FROM PREMISES.
@@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.
Thank you very much, your video was very helpful for me even if i am a college student!!
log base 5 (x) = log (x) ÷ log (5)
nice video!
but what do we do when constant part is equal to zero?
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; ∞).
are there other complex solutions?
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
I didn't get the note... how did that come!!
Is log base i a legitimate function? If so is it useful?
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)
Thank u!!! But how ro solve this? (x-1)Log3_(x)=x+1)/2
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?
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
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.
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.
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
thank you omg
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??
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
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
Abdulaziz Albaiz yeah i realized it after posting xD
great to see a fellow Arab on this channel bte
#logarithm #log #logbase5 #logbase7 #logbase25
A mathematician using base-10 logarithms?!
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!
The restriction of logs is that the base are different of 1, or i am wrong?
From the first look it's obviously x=1. It's so trivial!
without solving the equation , you can tell x=1 is the solution
I like blackpenredpen.
Thanks you
Just take the base as X and you will understand the problem and why it should be 1 intuitively.
well done!
thank you.
But Log_5(x+1)+log_7(x-1)=log_25(x)?
Can you solve this inequaliti please
log₃(x+2)
I think just making all of the logarith into log5 is way too ezier.
Can you help me with an interesting calculus problem? Its the integral of (x*lnx)^50 from 0 to 1.
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?
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.
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).
thanks
Ah yes, the loyarithm.
Wonderful
make log with base i
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
@@polyhistorphilomath you can check the answer in wolfram alpja
Tnx!
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🍻
What's with the easy questions man?
if we have different bases and different logarithms. how i can do? Like x, y, z.... help me please black pen red pen
why cant 1/log5 + 1/log7 -1/log25 equal 0??
can someone please explain??
thanks
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.
niroo s You could just put it in a calculator
The question is wrong because log to the base 1 is undefined.
Only at x=1
lifesaver
I hope can you help me in this E.Q. Log_3(x)+log_9(x)=6 thanks Soumyadip.
Log(X)=4 is the answer
@@spamspam9495 with a base of 3, right?
What the actual hell...
2:45 好像不小心要講出中文了
真的... hahah
15th
X is equal to Huaan xD
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.