I get x = 1 and x = e^2 for the question he gives us in the video (i.e. NOT the equation he solves). It is like a shorter version of the video equation.
I solved it by squaring both sides Then I made everything as a exponent of e So it was e^(ln√x)^2 = x Then I divided the (ln√x)^2 into two Put e^ln√x into a parenthesis and cancelled Then it simplifies to x = (√x)^ln√x I squared both sides to get x^2 = x^ln√x Took the log base x on both sides Then I got ln√x = 2 I made everything as exponent of e On the left side I got √x = e^2 Squared both sides And got x = e^4
Note: When you typed e^(ln√x)^2 it's best if you instead type e^[(ln√x)^2] in brackets or parenthesis to show that (ln√x)^2 is the entire exponent of e. Your method is partially correct, though it's astonishingly complicated. One of the main things to pursue in maths is simplicity, especially as a student. This will really help you in understanding and getting used to maths. In the case of algebra, in most equations like this you should always attempt to bring everything to one side so it all equals to zero, then simplify as much as possible and use simple methods such as factoring and the quadratic formula. I called your method partially correct because it only found one of two solutions. I hope my comment was helpful!
You can actually manipulate the first equation to get the last one. Take u^2 to be x, and subbing it in you get ln(sqrt(u^2))=sqrt(ln(u^2)). Cancelling the square and square root on the left and then squaring both sides gives, (lnu)^2=ln(u^2). Since u is in terms of x, we can say that u=sqrt(x) taking only positive because of the way we canceled the square and square root.
For the question at the end of the video, the answer is x = 1 or x = e**2. The key is noticing that ln(x**2) = 2.ln(x). If ln(x) = 0 (x = 1), then the egality is already satisfied. Else, by simplifying, we'll have 2 = ln(x), so that x is equal to e**2.
2 ปีที่แล้ว +1
Below is for the question answered in the video. The question left to the viewer is "the opposite" so x=1 or x=e^2. I wasn't expecting that question. Very easy, but very nice "dual" problem anyway.
Under the condition of x>0, we can bring the exponent of x as a multiple of the logarithm on the left side, move it to the right and factor out ln(x) and then we have a similar situation. On both occasions we can exponentiate and end up with x=1 or x=e^2 which are both positive so they meet the criteria.
if you want the equation to be ln(x^2)=(lnx)^3 it's easily solvable too. That means (lnx)^3-2lnx=0, after factorizing lnx can be 0, sqrt(2), -sqrt(2). So x is either 1, e^sqrt(2), 1/(e^sqrt(2)).
For x>0 and "a" a real number, if a=1 or x=1, then the egality is already satisfied. Otherwise, we can simplify by ln(x) since x is not 1. We will have a=ln(x)**(a-1), and because a-1 is not equal to 0, we'll have ln(x)=a**(1/(a-1)), to reach out to the formula you wrote. In general, the case a=1 will make the egality true for all x, otherwise we'll have the formula you wrote.
If a=0, then the exponential formula is equal to exp(0) = 1. We return to the 1st case where x = 1. And since x > 0, then the exponential formula is right even if a 0 or < 0, what matters is that a should be not equal to 1 in the exp formula. If "a=1", then ln(x^1) = ln(x)^1 which is already true for all x>0, there is nothing to prove in the case a=1.
ln(x^2) = (lnx)^2 2lnx = (lnx)^2 0 = (lnx)^2 - 2lnx (lnx)^2 - 2lnx = 0 lnx(lnx - 2) = 0 lnx = 0 x = 1[log base a of 1 equals 0, where a is not equal to 0] lnx - 2 = 0 lnx = 2 log base e of x = 2 e^2 = x Therefore, x = 1, e^2 It took me a while to find this path and it wasn’t so hard to be honest
For the last problem (final written) [ ln(x^2)=(ln x)^2 ]: My answer is x = 1, e^2 At first, from the right side, x>0. ∴ In(x^2) = 2*ln |x| = 2(ln x) (∵ x>0) By setting " ln x = t ", 2t = t^2 . ∴ t^2-2t=0 ∴ t(t-2)=0 ∴ t=0 or t=2 ∴ ln x = 0, 2 ∴ x = 1 (from " ln x = 0" ), e^2 (from " ln x = 2 " ) [Check: ] x=1 Left side: ln(1^2)=ln(1)=0, Right side: (ln 1)^2=0^2=0 OK. x=e^2 Left side: ln((e^2)^2)=ln(e^4)=4×ln(e)=4×1=4, Right side: (ln(e^2))^2=(2×ln(e))^2=2^2=4 OK. By the way, are there any videos that show the answers of the homework for the last problem? I commented with my answer about another video in the past. But, I have never found the videos about "homework answer" for this kind of video. So, I asked about it. Anyway, after doing homework, viewers cannot get to know correct answers. For a lot of educational you-tube videos, they have same problems. Concerning me, I made and upload my videos about math problems on you-tube, too. For me, I sometimes introduce homework. Then, I will make "homework answer video" for the next video. Ref (please watch) : th-cam.com/channels/ahi9I-OtJ3xPKfLiFUMqAQ.html [Note: main language is Japanese] But, a lot of "math problem you-tubers" do Not try to do like I do. Why? Appendix: For the last problem (first written before correcting) [ ln(x^2)=(ln x)^3 ]: My answer is x = 1, e^(√2), and e^(-√2) [= 1/(e^(√2)) ] At first, from the right side, x>0. ∴ In(x^2) = 2*ln |x| = 2(ln x) (∵ x>0) By setting " ln x = t ", 2t = t^3 . ∴ t^3-2t=0 ∴ t(t^2-2)=0 ∴ t = 0 or t^2=2 ∴ t = 0, ±√ 2 ∴ ln x = 0, , ±√ 2 ∴ x = 1 (from " ln x = 0" ), e^(√2) (from " ln x = √2" ), e^(-√2) (from " ln x = -√2" ) [Check: ] x=1 Left side: ln(1^2)=ln(1)=0, Right side: (ln 1)^3=0^3=0 OK. x=e^(√2) Left side: ln((e^(√2) ^2)=ln(e^(2√2))=2√2×ln(e)=2√2×1=2√2, Right side: (ln(e^(√2))^3=((√2)×ln(e))^3=(√2)^3=2√2 OK. x=e^(-√2) Left side: ln((e^(-√2)^2)=ln(e^(-2√2))=(-2√2)×ln(e)=(-2√2)×1=-2√2, Right side: (ln(e^(-√2))^3=((-√2)×ln(e))^3=(-√2)^3=-2√2 OK.
Actually, you can solve it like a quadratic equation: we can write ln√x = √(lnx) as e^A = √x and e^A² = x, where A is some number. By taking the square root of the last equation, we get e^(A²/2) = √x [Because the square root is basically n^(1/2)]. And since e^A also equals √x, we can say that A = A²/2, A² = 2A. And now we can just write this as a quadratic equation: A² - 2A + 0 = 0. I assume you know how to solve the rest :)
So many people are getting x = 1 and e². How? All I can see is, somehow, they're putting sqrt(x) as x. sqrt(x) = 1 and e², i.e. x = 1 and e⁴. ln(sqrt(e²)) = ln(e¹) = lne = 1 But sqrt(ln(e²)) = sqrt(2) which is not 1. However, ln(sqrt(e⁴)) = ln(e²) = 2, and sqrt(ln(e⁴)) = sqrt(4) = 2.
You forgot the 3rd option, which is x=1, because for x>0 we have : ln(x^2)=ln(x)^3 => 2*ln(x) = ln(x)^3 => ln(x)[2 - ln(x)^2]=0 => ln(x)=0 or ln(x)=+-sqrt(2) => x=1 or x=e^(+-sqrt(2))
How ironic, I got e^4, but didn't find out 1, because I divided the ln(x) instead of substracting it. So I had 1/4 * ln(x) = 1, which only result would be e^4. Why is it like that? Can someone explain? But for the question at the end I got: x=1, x=e^2
When you divide by ln x, you are excluding ln x = 0 as a possible answer (because you can't divide by zero). That's why factorization is necessary: to not exlude any possible answers.
No !! We can't. Let's suppose by absurd that we can. We'll get : exp(sqrt(x))=exp(x/2) Meaning that : sqrt(x)=x/2 => 2*sqrt(x) = x => x=0 or sqrt(x) = 2 => x=0 or x=4 In the "ln" case we have 1 and e**4 as the solutions, and in the "exp" case we have 0 and 4 as the solutions. Which is a contradiction. So in general we should not replace a function with any other function in an equation. We can turn the ln into exp if we have an equation of the form ln(f(x)) = ln(g(x)), take out the ln function since it's injective and enter the exp function, but in the equation we have sqrt(ln(f(x))), so we can't do always the transformation you mentioned.
The basic idea of log is it expresses a number as a power of the given base. In practical terms, to find the log of a number to a given base means answering the question "what power?". For example, log to base 10 of 100 (written log 100) is 2, as we must say what power of 10 gives 100, and 100 equals 10 to the power of 2, i.e. 100=10². Similarly, log to the base 2 of 8 (written log₂8) is 3, as 8=2³. In general, if b=aˣ then logₐb=x and vice versa. Technically, log to the base a of x (where a>0 and a≠1) is the inverse function of a^x. When we write log without a base we generally mean base 10 (except in university maths) and ln means log to the base e (Euler's number). From this basic idea we derive the rules of logs from the rules of exponents, for example log xy=log x + log y (where the logs are all to the same base) comes from aˣ⁺ʸ=aˣaʸ. You can find these rules in any suitable textbook. I hope this gives you some idea of logarithms.
You forgot the 3rd option, which is exp(-sqrt(2)). Now x = 1 is already a solution. Otherwise, by simplifying with ln(x), we'll have ln(x)**2 = 2. It implies that ln(x) = sqrt(2) or - sqrt(2), since ln(x) could take negative values. We can finally conclude that x = e**(sqrt(2)) or x = e**(-sqrt(2)).
@@fahdfarachi6232 I don't know that ln(x) can take neg value. I didn't study that in school (just prelearn it though videos on his channel) so when I saw he write sqrt(x) > 0, I thought that ln(x) can only take positive value. Thank you for showing me my mistake :)
I get x = 1 and x = e^2 for the question he gives us in the video (i.e. NOT the equation he solves). It is like a shorter version of the video equation.
e^4 not e^2
@@2012tulio yes, for the video question. I answered his question at the end of the video: ln(x^2) = (ln(x))^2
i got x=1 and x=e^2^1/2(e power root 2)
I get x = 1 and x = e^2
@@arulvel96 what was your method?
I solved it by squaring both sides
Then I made everything as a exponent of e
So it was e^(ln√x)^2 = x
Then I divided the (ln√x)^2 into two
Put e^ln√x into a parenthesis and cancelled
Then it simplifies to x = (√x)^ln√x
I squared both sides to get x^2 = x^ln√x
Took the log base x on both sides
Then I got ln√x = 2
I made everything as exponent of e
On the left side I got √x = e^2
Squared both sides
And got x = e^4
Note: When you typed e^(ln√x)^2 it's best if you instead type e^[(ln√x)^2] in brackets or parenthesis to show that (ln√x)^2 is the entire exponent of e.
Your method is partially correct, though it's astonishingly complicated. One of the main things to pursue in maths is simplicity, especially as a student. This will really help you in understanding and getting used to maths. In the case of algebra, in most equations like this you should always attempt to bring everything to one side so it all equals to zero, then simplify as much as possible and use simple methods such as factoring and the quadratic formula.
I called your method partially correct because it only found one of two solutions. I hope my comment was helpful!
Since ln(x²)=(lnx)² is an assignment, I'll submit it tomorrow as a video 😁
You can actually manipulate the first equation to get the last one. Take u^2 to be x, and subbing it in you get ln(sqrt(u^2))=sqrt(ln(u^2)). Cancelling the square and square root on the left and then squaring both sides gives, (lnu)^2=ln(u^2). Since u is in terms of x, we can say that u=sqrt(x) taking only positive because of the way we canceled the square and square root.
Q: x=1 , e^2
السلام عليكم يا صديقي العربي
For the question at the end of the video, the answer is x = 1 or x = e**2. The key is noticing that ln(x**2) = 2.ln(x). If ln(x) = 0 (x = 1), then the egality is already satisfied. Else, by simplifying, we'll have 2 = ln(x), so that x is equal to e**2.
Below is for the question answered in the video. The question left to the viewer is "the opposite" so x=1 or x=e^2. I wasn't expecting that question. Very easy, but very nice "dual" problem anyway.
ln(x^2)=(ln(x))^3
x=1
x=e^(-sqrt(2))
x=e^(sqrt(2))
Cool!!!!😊
Thanks so much blackpenredpenbluepengreenpen!!!
Under the condition of x>0, we can bring the exponent of x as a multiple of the logarithm on the left side, move it to the right and factor out ln(x) and then we have a similar situation. On both occasions we can exponentiate and end up with x=1 or x=e^2 which are both positive so they meet the criteria.
if you want the equation to be ln(x^2)=(lnx)^3 it's easily solvable too.
That means (lnx)^3-2lnx=0, after factorizing lnx can be 0, sqrt(2), -sqrt(2).
So x is either 1, e^sqrt(2), 1/(e^sqrt(2)).
To generalize: ln(x^a)=(ln x)^a has solutions x=1 and x=e^a^(1/(a-1)). But what is the domain of a?
For x>0 and "a" a real number, if a=1 or x=1, then the egality is already satisfied. Otherwise, we can simplify by ln(x) since x is not 1. We will have a=ln(x)**(a-1), and because a-1 is not equal to 0, we'll have ln(x)=a**(1/(a-1)), to reach out to the formula you wrote.
In general, the case a=1 will make the egality true for all x, otherwise we'll have the formula you wrote.
@@fahdfarachi6232 So you're saying that "a" can be any real number except 1 to reach that formula?
@@PlutoTheSecond Yes, because if a=1, then the egality is true for all x>0. But "a is not 1" is necessary to reach the exponential formula.
@@fahdfarachi6232 You might want to think about this some more. What happens if a=0? What happens if a
If a=0, then the exponential formula is equal to exp(0) = 1. We return to the 1st case where x = 1. And since x > 0, then the exponential formula is right even if a 0 or < 0, what matters is that a should be not equal to 1 in the exp formula. If "a=1", then ln(x^1) = ln(x)^1 which is already true for all x>0, there is nothing to prove in the case a=1.
ln(x^2) = (lnx)^2
2lnx = (lnx)^2
0 = (lnx)^2 - 2lnx
(lnx)^2 - 2lnx = 0
lnx(lnx - 2) = 0
lnx = 0
x = 1[log base a of 1 equals 0, where a is not equal to 0]
lnx - 2 = 0
lnx = 2
log base e of x = 2
e^2 = x
Therefore, x = 1, e^2
It took me a while to find this path and it wasn’t so hard to be honest
For the last problem (final written) [ ln(x^2)=(ln x)^2 ]: My answer is x = 1, e^2
At first, from the right side, x>0. ∴ In(x^2) = 2*ln |x| = 2(ln x) (∵ x>0)
By setting " ln x = t ", 2t = t^2 . ∴ t^2-2t=0 ∴ t(t-2)=0 ∴ t=0 or t=2 ∴ ln x = 0, 2
∴ x = 1 (from " ln x = 0" ), e^2 (from " ln x = 2 " )
[Check: ] x=1 Left side: ln(1^2)=ln(1)=0, Right side: (ln 1)^2=0^2=0 OK.
x=e^2 Left side: ln((e^2)^2)=ln(e^4)=4×ln(e)=4×1=4, Right side: (ln(e^2))^2=(2×ln(e))^2=2^2=4 OK.
By the way, are there any videos that show the answers of the homework for the last problem?
I commented with my answer about another video in the past. But, I have never found the videos about "homework answer" for this kind of video.
So, I asked about it. Anyway, after doing homework, viewers cannot get to know correct answers. For a lot of educational you-tube videos, they have same problems.
Concerning me, I made and upload my videos about math problems on you-tube, too. For me, I sometimes introduce homework. Then, I will make "homework answer video" for the next video. Ref (please watch) : th-cam.com/channels/ahi9I-OtJ3xPKfLiFUMqAQ.html [Note: main language is Japanese]
But, a lot of "math problem you-tubers" do Not try to do like I do. Why?
Appendix:
For the last problem (first written before correcting) [ ln(x^2)=(ln x)^3 ]: My answer is x = 1, e^(√2), and e^(-√2) [= 1/(e^(√2)) ]
At first, from the right side, x>0. ∴ In(x^2) = 2*ln |x| = 2(ln x) (∵ x>0)
By setting " ln x = t ", 2t = t^3 . ∴ t^3-2t=0 ∴ t(t^2-2)=0 ∴ t = 0 or t^2=2 ∴ t = 0, ±√ 2 ∴ ln x = 0, , ±√ 2
∴ x = 1 (from " ln x = 0" ), e^(√2) (from " ln x = √2" ), e^(-√2) (from " ln x = -√2" )
[Check: ] x=1 Left side: ln(1^2)=ln(1)=0, Right side: (ln 1)^3=0^3=0 OK.
x=e^(√2) Left side: ln((e^(√2) ^2)=ln(e^(2√2))=2√2×ln(e)=2√2×1=2√2, Right side: (ln(e^(√2))^3=((√2)×ln(e))^3=(√2)^3=2√2 OK.
x=e^(-√2) Left side: ln((e^(-√2)^2)=ln(e^(-2√2))=(-2√2)×ln(e)=(-2√2)×1=-2√2, Right side: (ln(e^(-√2))^3=((-√2)×ln(e))^3=(-√2)^3=-2√2 OK.
Man put an appendix 💀
Actually, you can solve it like a quadratic equation: we can write ln√x = √(lnx) as e^A = √x and e^A² = x, where A is some number. By taking the square root of the last equation, we get e^(A²/2) = √x [Because the square root is basically n^(1/2)]. And since e^A also equals √x, we can say that A = A²/2, A² = 2A. And now we can just write this as a quadratic equation: A² - 2A + 0 = 0. I assume you know how to solve the rest :)
2ln x = (ln x)²
lnx(lnx-2)=0
x=1 or x=e²
So many people are getting x = 1 and e². How? All I can see is, somehow, they're putting sqrt(x) as x.
sqrt(x) = 1 and e², i.e. x = 1 and e⁴.
ln(sqrt(e²)) = ln(e¹) = lne = 1
But sqrt(ln(e²)) = sqrt(2) which is not 1.
However, ln(sqrt(e⁴)) = ln(e²) = 2, and
sqrt(ln(e⁴)) = sqrt(4) = 2.
People are solving the equation set right at the end of the video, ln(x²)=(ln x)², not the original equation, ln(√x)=√(ln x).
Q:lnx^2=(lnx)^2
x=1; x=e^2. Is my solution right?
Results are x=1 and x=e^2
ln(sqrt(x)) = sqrt(ln(x))
1/2ln(x) = sqrt(ln(x))
(y := ln(x))
1/2y = sqrt(y)
1/4y^2 = y
1/4y^2 - y = 0
y1,2 = (0,4)
x1 = 1
x2 = e^4
The solutions are: x = 1, e^4.
For ln(x^2) = (lnx)^3 gives ln(x) = +/-√2 does give; x = e^+√2 or x = e^-√2 not sure though
You forgot the 3rd option, which is x=1, because for x>0 we have :
ln(x^2)=ln(x)^3 => 2*ln(x) = ln(x)^3
=> ln(x)[2 - ln(x)^2]=0
=> ln(x)=0 or ln(x)=+-sqrt(2)
=> x=1 or x=e^(+-sqrt(2))
Thank you Teacher ❤️
Let y=ln(x), that equation can change into: 1/2*y=y^(1/2)
Yes I finally figured one out on my own!
I did it in my head in a minute. x=1 and x=e^4
Now watching the video...
2lnx=ln²x
lnx=u
2u=u²
u²-2u=0
u=0 or u=2
x=1 or x=e²
If (sin x)^2 can be written as sin^2 (x), why can't (ln x)^2 be written as ln^2 (x)?
Some people do that, mainly if they are lazy.
@@MusicalInquisit as a lazy person I can confirm
I do write it like that
e^2/1
3:11 i would substitute ln(x) with a
ln(x²) = ln²x
D: x² > 0 and x > 0
D: x > 0
ln²x - 2lnx = 0
lnx * (lnx - 2) = 0
(1) lnx = 0 or (2) lnx - 2 = 0
(1) lnx = 0, x = 1
(2) lnx - 2 = 0, lnx = 2, x = e²
Answer: 1; e²
For the ln^2 question, my answer is e^2. And 1 ***
How ironic, I got e^4, but didn't find out 1, because I divided the ln(x) instead of substracting it.
So I had 1/4 * ln(x) = 1, which only result would be e^4. Why is it like that? Can someone explain?
But for the question at the end I got:
x=1, x=e^2
When you divide by ln x, you are excluding ln x = 0 as a possible answer (because you can't divide by zero). That's why factorization is necessary: to not exlude any possible answers.
Nice equation 👍
Does 0 count?
The 3s in the backkkk 🤘
x = 1, x = e^2
What "in" mean?
The asnwer is 1 and e^4
lnX=2 = e²
e² 1
x=1
x=e^4
x=1 or x=e^4
X=1 and e^2 too easy blackpenredpen.
can you turn the ln to exp and solve it🤔
No !! We can't.
Let's suppose by absurd that we can.
We'll get : exp(sqrt(x))=exp(x/2)
Meaning that :
sqrt(x)=x/2 => 2*sqrt(x) = x
=> x=0 or sqrt(x) = 2
=> x=0 or x=4
In the "ln" case we have 1 and e**4 as the solutions, and in the "exp" case we have 0 and 4 as the solutions.
Which is a contradiction.
So in general we should not replace a function with any other function in an equation.
We can turn the ln into exp if we have an equation of the form
ln(f(x)) = ln(g(x)), take out the ln function since it's injective and enter the exp function, but in the equation we have sqrt(ln(f(x))), so we can't do always the transformation you mentioned.
!a very good video like alway
The answer of new question is 1 and e^2
i got x = 1 and x = exp(2)
This was really easy
What the jordan doing back there
I don't understand logarithms ;_;
Pleas sb explain me
The basic idea of log is it expresses a number as a power of the given base.
In practical terms, to find the log of a number to a given base means answering the question "what power?".
For example, log to base 10 of 100 (written log 100) is 2, as we must say what power of 10 gives 100, and 100 equals 10 to the power of 2, i.e. 100=10².
Similarly, log to the base 2 of 8 (written log₂8) is 3, as 8=2³.
In general, if b=aˣ then logₐb=x and vice versa.
Technically, log to the base a of x (where a>0 and a≠1) is the inverse function of a^x.
When we write log without a base we generally mean base 10 (except in university maths) and ln means log to the base e (Euler's number).
From this basic idea we derive the rules of logs from the rules of exponents, for example log xy=log x + log y (where the logs are all to the same base) comes from aˣ⁺ʸ=aˣaʸ. You can find these rules in any suitable textbook.
I hope this gives you some idea of logarithms.
good equation m.A.A.d identity
If it is ln(x²) = (lnx)³
x = 1, x = e^sqrt(2)
You forgot the 3rd option, which is exp(-sqrt(2)). Now x = 1 is already a solution. Otherwise, by simplifying with ln(x), we'll have ln(x)**2 = 2. It implies that ln(x) = sqrt(2) or - sqrt(2), since ln(x) could take negative values. We can finally conclude that x = e**(sqrt(2)) or
x = e**(-sqrt(2)).
@@fahdfarachi6232 I don't know that ln(x) can take neg value. I didn't study that in school (just prelearn it though videos on his channel) so when I saw he write sqrt(x) > 0, I thought that ln(x) can only take positive value. Thank you for showing me my mistake :)
@@ntth74 You're welcome
That was a tough one!