วีดีโอ

ln(50)/ln(5)

Решаем методом устного счета. 144=12^2 X=12^1/6 Good luck!

Germany? An utter insult to Gauss, Riemann, Cantor, HIlbert and many others. I pray that nobody taking complex analysis for the first time comes across this.

10¹⁰ * 10¹⁰ ---------------- = (10÷2)¹⁰ = 5¹⁰ 10¹⁰ * 2¹⁰ So what was the "challenge" in your problem ? 😄🤪

This video appears to be a math troll. In general, e^(ln(z)) = z always but ln(e^(z)) ≠ z This happens because w=ln(z) if, and only if, e^(w) = z And, for any integer k, e^(w+i2kπ) = e^(w) (See [a] at the end) Therefore there are infinite values of the complex logarithm (it's a many-valued function) and the equality makes no sense. To make the logarithm a plain old function we choose k so that the imaginary part of the logarithm is greater than -π and less than or equal to π. This is the so called "principal value". (The same thing happens to the complex root operator, that yields the principal value of the root and can be misused to make a video similar to this one...) So in the complex numbers ln(z^x) ≠ x ln(z) in general. And also taking logarithm in both sides of an equation adds infinite superfluous solutions (if we don't take the principal value) and, for the principal value, only works if the imaginary part of both sides is greater than -π and less than or equal to π. [a]: e^(w+i2kπ) = (e^w)(e^(i2kπ)) = (e^w)(cos(2kπ)+i sin(2kπ)) = (e^w)(1 + i0) = e^w

1^1/1^2 1/2 (x ➖ 2x+1).

Why not change-8 to (-2)^3. That way you get (x -2)^3 = (-2)^3 . Get rid of the powers since both sides are raised to the power of three so you get x -2 = -2. Add 2 to both sides so x = 0?

0

I don't like the mixed unrelated videos.

I just looked at this for a few seconds and got the answer. 4 to the power of 1/4 (0.25) is approximately 1.41421 which is the value of X. Raise that number to the power of four equals four. I agree with previous comment. Stop click baiting.

Can you make the exponent font any smaller???

0.5

1x1x1x1x1x1x ..... = 1 every and 1≠2 ...

I instantly find x is equal to +- sq rt 2. Stop clickbaiting

Dude, you're teaching very poorly. The unit factor raised to anything is always equal to the unit factor! If that weren't the case, I could easily prove that ln 2 = 0 and also that 2 = 1. Well, and in every bank in the world they'll explain to you that 2 can't equal 1, because they'd go bankrupt. Try some art.

Estou aprendendo muito aqui no Brasil.

x=(log7-log2)/log5 x≈(0.845-0.301)/0.699≈544/699≈0.7783 Deviation is about 0.0001 2×5^0.7783≈6.999

isn't k = 0 ? if it is we are still dividing by 0

x=log128/log4=log2⁷/log2²=7log2/2log2=7/2=3.5

x=log36/log9=(log9+log4)/log9=1+log2/log3 x≈1+0.301/0.477≈778/477≈1.631 Deviation is about 0.0001 9^1.631≈36.0056

X-1.64

x = [1, -1, i, -i, 0] Am I a genius ? 🤪😁 x⁵ - x³ = 0 x³(x² - 1) = 0 x³ = 0, x1 = 0 obviously x² = 1 x2 = 1, x3 = -1

X-2.47

Wikipedia: Leonhard Euler (/ˈɔɪlər/ OY-lər;[b] German: [ˈleːɔnhaʁt ˈʔɔʏlɐ] ⓘ, Swiss Standard German: [ˈleɔnhard ˈɔʏlər]; 15 April 1707 - 18 September 1783) was a Swiss polymath who was active as a mathematician, physicist, astronomer, logician, geographer, and engineer. He founded the studies of graph theory and topology and made influential discoveries in many other branches of mathematics such as analytic number theory, complex analysis, and infinitesimal calculus. He also introduced much of modern mathematical terminology and notation, including the notion of a mathematical function.[6] He is also known for his work in mechanics, fluid dynamics, optics, astronomy, and music theory.[7] As a result, Euler has been described as a "universal genius" who "was fully equipped with almost unlimited powers of imagination, intellectual gifts and extraordinary memory". Doctor of Law Doctor of Medicine also.

i'm a 12th grade student and we haven't reached the log yet, but for the second equation we also the ln right?

You have at least 20 of these problems and it certainly does not take a genius to solve these. Your average 9th grader could solve them easily. I wish it was possible to block your channel.

In the very first problem you forgot the 1 so the answer would have been 1 + log base 3 of 2

Hey! Hello, I tried to resolve this question too, my answer was this: 9ⁿ = 36 First I did the following property: a⁴ = (a²)² (3²)ⁿ = 36 Later I multiplied the exponent of the base with the exponent outside the parentheses giving the result as: 3²ⁿ = 36 With that, I noticed that I could use log₃ to cancel with the logarithm, that is, 3²ⁿ Applying log₃ on both sides, it will be as follows: log₃(3)²ⁿ = log₃(36) Therefore the first member of the equality will be as: 2n = log₃(36) and with that the answer will be: n = log₃(36) / 2

I got x = 1.187

ln2/2пni, n is integer, not equal to 0.

Literally not possible because 1 powered by any finite number is 1

Are you Black?

Answer x -2

9:02 log 10 is not 1 Its 1 in chemistry only where base is taken as 10 In mathematics log x is a logaritm with natural bade which is e(or eular's number)

Yes I can.

Okay, chat.. Where is this wrong?

lmaoo the beeping sound

Do need to change the battery in your smoke alarm? Lol

This beeping makes the Vid unwatchable ...

(1)^z = {e^(2kπi +ln(1)}^z = {e^(2kπi +0}^z = e^(2kπi.z) with k∈ ℤ. So (1)^z = 2 = e^(ln2) implies e^(2kπi.z) = e^(ln2) with k ≠ 0 bec. e^0=1≠2. Thus 2kπi.z = ln2 and so z = (ln2)/(2kπi) = - i.(ln2)/(2kπ) for k ≠ 0. ∴ there are an infinite number of roots, one for each value of k∈ ℤ- {0}.

W vid helped me for my math exam.

Thank you for this video. It has me confused which leads to questions. when i first completed the problem, I got X = 2 + log2 base 5. is this answer correct?

I must be a 80 years old genius, it took me 10 secs to solve.

Why would you waste 11+ minutes on this stupidity. Use a calculator.

Solution: 4^x+6^x = 9^x ⟹ 2^(2x)+2^x*3^x = 3^(2x) |/[2^x*3^x] ⟹ 2^x*2^x/[2^x*3^x]+1 = 3^x*3^x/[2^x*3^x] ⟹ (2/3)^x+1 = (3/2)^x ⟹ (2/3)^x+1 = 1/(2/3)^x |with u = (2/3)^x ⟹ u+1 = 1/u |*u ⟹ u²+u = 1 |-1 ⟹ u²+u-1 = 0 |p-q-formula ⟹ u1/2 = -1/2±√(1/4+1) = -1/2±1/2*√5 = (-1±√5)/2 ⟹ u1 = (-1+√5)/2 = (√5-1)/2 and u2 = -(1+√5)/2 ⟹ 1st case: (2/3)^x1 = u1 = (-1+√5)/2 = (√5-1)/2 |ln() ⟹ x1*ln(2/3) = ln[(√5-1)/2] |/ln(2/3) ⟹ x1 = ln[(√5-1)/2]/ln(2/3) ≈ 1,1868 2nd case: (2/3)^x2 = u2 = -(1+√5)/2 < 0 [that is not defined]

skibidi Rizz

Pretty sure Im not a genius, and I solved it on the toilet. Do I get a medal?

simple x = 1,0 or -1

4^1+6^1=9^1?

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