Solving An Interesting Exponential Equation

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[@gregwochlik9233]

x = 8 by guessing at 0:08. I recognise 1 + i as sqrt(2).e^i(pi/4). Squaring it gives 2i. Squaring again gives -4. Squaring again gives 16

[@bugsfudd8295]

Me too! Good job!

[@BlaqRaq]

I can hardly multiply in forward and you want me to multiply in reverse?!?

[@SyberMath]

Sure. Why not? 😁

[@BlaqRaq]

@@SyberMath 🤣

[@sumit180288]

Using polar form will be a much simpler method to solve this problem.
We will get x = 8 without doing all this circus solution by polar form method.

[@richardmayerthaler3645]

simple solution: 1 + i = SQRT 2 ..... SQRT2 ^8 = 16 ....... or 1 + i = 16 ^ 1/x ...... SQRT 2 = 16 ^ 1/x ... thats much more simple

[@vladyslavboboshko1566]

"We gonna be solving for the X-value' - Do you have any alternative?))

[@SyberMath]

that makes me sad :(
jk 😜😁😍

[@rickdesper]

Use polar form. r = sqrt(2), theta = pi/4. r^x = 16, theta*x = 2n*pi for some integer n. x = 8.

[@carly09et]

|1 + i| = root(2) |16| = 16 which is [root(2)]^8
angle is pi/4 8.pi/4 = 2pi so 8 is the real solution
the complex solution is as the first method, I can't think of a short cut to get it :(
a == root(2) . e^i[pi/4]
a^x == [root(2)]^x . e^(i[pi/4].x) == 16

[@andreybyl]

Open the textbook on complex analysis. This is an invalid entry. The left side of the equation is a set and the right side is a number. and you can't take the logarithm of both sides of the «equation», the complex logarithm is a multivalued function. Secondly, again, it is impossible to take the logarithm of the set, but even if we take any one branch of the logarithm Ln(z) and denote it ln(z) and use it to determine the complex powers z ^ w, then it is not true that ln (z ^ w) =w*ln(z)))) Your entire solution is completely incorrect and wrong ((
Example: If ln(z):=ln|z|+i(arg(z) + 2pi) , and z^w := exp(w*ln(z)) then ln( 1^sqrt(2)=/ =sqrt(2)*ln(1)

[@alextang4688]

I think polar form can solve the problem. 😉😉😉😉😉😉

[@rickdesper]

Indeed. I can see no reason to use another method.

[@grumpyentity2029]

How do you show, by subbing back into the original equation, that the solution works? The amount of algebra is ridiculous.

[@RexxSchneider]

Before watching: On the LHS, (1+i) has a modulus of √2, so if we're going to match the modulus of 16 (=2^4) on the RHS, we need to raise (1+i) to the power 8. So it looks like x = 8.
Check: (1+i)^8 = (√2)^8 * e^(πi/4)^8 = 16 * e^(2πi) = 16 * 1 = 16.
After watching: I was surprised by the general solution, as I was expecting only one. By inspection of the general solution, I think that the only way that the imaginary terms can disappear, is if k and n are chosen to make the numerator a real multiple of the denominator, and so in order to cancel the ln2, that multiple must be 8, making x=8 the only real solution. Thoughts?

[@SyberMath]

Exactly!

[@drdiegocolombo]

Is it really necessary, for the resolution, to transform 16 into 16 e^i2kπ?.

[@SyberMath]

To find complex solutions, yes

[@ytlongbeach]

i appreciate the lack of rigor on the 2nd method, for the win 1!!11

[@SyberMath]

Yay for the no-rigor team!!! 😜😍

[@PunmasterSTP]

This was exponentiawesome! 👍

[@SyberMath]

That's a ginormous complement! Thanks a million! 😉🧡

[@scottleung9587]

The second method was awesome!

[@SyberMath]

nice and clean, right? 😍

[@SIB1963]

I enjoy your vids.

[@SyberMath]

Glad to hear that 🧡

[@vladimirkaplun5774]

Huge number of 10 mins tasks. For example (1+i√3)^x=64 and so on...😁

[@SyberMath]

I try to keep it short but you know I keep talking and get carried away 😜😁

[@ilyashick3178]

x=4/log_2 of (1+i)

[@giuseppemalaguti435]

x=ln16/ln(1+i)