I Solved An Interesting Exponential Equation

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[@mcwulf25]

I first divided by 25 to get an exponent (x-2) on the left and -1 on the right.
Then proceed as you did, subtracting 2 at the end.

[@angeluomo]

Brilliant. Makes the method I proposed even simpler.

[@angeluomo]

Great video! How about this approach? 5^x=-25 -> 5^x = 25*-1 -> xln5=ln25 + ln(-1) -> x=ln25/ln5 + ln(-1)/ln(5) -> x = 2 + (pi*i)/ln5 (since pi*i = ln(-1)). Then pi*i can be expressed as (2n+1)pi*i.

[@scottleung9587]

Got it!

[@barakathaider6333]

👍

[@tamilselvanrascal5956]

🎉

[@levskomorovsky1762]

x ln 5 = ln (-25)
x ln 5 = ln 25 + ln (-1)
x ln5 = 2 ln 5 + ln (e^[2n +1)i π]
x = 2 +[ 2n +1)i π]/ln5

[@Christian_Martel]

Wow, I didn’t thought of this shortcut!! Thanks!

[@allanmarder456]

Sorry everyone, I made a mistake. My error was when I wrote 5^x = (i5)^2. That gives only part of the answer. I can also write 5^x=(-i5)^2. When you solve this
you get the missing part and taken together the answer in the video is indeed correct. Sorry for the confusion. WHAT FOLLOWS IS ONLY HALF RIGHT!!
I'm getting a slightly different answer. First i= e^i(pi)/2 and 1=e^2(pi)n*i n an integer. ln(i)=ln(e^(i(pi)/2 + 2(pi)n*i) = i(pi)/2 + 2(pi)n*i = 2*(pi)i * (1/4+n)
5^x =(i5)^2 Taking ln of both sides: x*ln(5) = 2*(ln(i) +ln(5)) x=2*ln(i)/ln(5) +2 = 2*[ 2*(pi)i * (1/4+n)]/ln(5) +2 =4*(pi)i*(1/4+n)/ln(5) +2 =(pi) [1+4n]/ln(5) +2

[@mcwulf25]

Interesting. Looks like by working with i rather than-1 then adding 2npi then squaring the expression, the periodicity becomes 4pi. Perhaps this is because you also need to take into account that -i also squares to -1, and you need to include (2n + 3/2) pi.i in your formulation?

[@mcwulf25]

Or put differently, 5^x = (-5i)^2 too.

[@allanmarder456]

@@mcwulf25Absolutely correct. I blew it! Thanks for the comment

[@foxyone3]

These are nice puzzles, could you relate them back to the solution of a real problem?

[@SyberMath]

Good question!

[@Ricardo_S]

5^x=-25
Take logbase5
x=log5(-25)
log of a negative is complex
x=log5(-1)+log5(25)
remember the change of base for logarithm
(ln(-1))/ln(5)+log5(5²)
-1=e^(i(π+2πk))
((ln(e^(i(π+2πk))))/(ln(5)))+2
(i((π+2πk))/(ln(5)))+2
k its an integer
x=log5(-25)=((i(π+2πk))/(ln(5)))+2

[@Qermaq]

This is making me nuts. Maybe someone can shed light.
The natural numbers : n
The triangular numbers: (n)(n+1)/2
The square numbers: n^2
The pentagonal numbers: essentially, these are the square of n plus the triangle of n minus the integer n.
Hexagonals are just pentagonal plus square minus triangle. Heptagonals are just hex plus pent minus square. And so on.
So we can go back a step. If pent(n) = n^2 + (n)(n+1)/2 - n, or pent(n) = sq(n) + tri(n) - int(n) then whatever is below int(n) - let's call it ?(n) = will follow the same pattern as all polygonal numbers above. So sq(n) = tri(n) + int(n) - ?(n), or ?(n) = int(n) + tri(n) - sq(n). So ?(n) = n + (n)(n+1)/2 - n^2.
This yields a very interesting sequence. Where do I learn more about this?

[@SyberMath]

I'll share two links:
qcweb.qc.edu.hk/math/Junior%20Secondary/Polygon%20number.htm
en.wikipedia.org/wiki/Polygonal_number

[@Qermaq]

@@SyberMath Will check out, thanks. I realized later that if you invert the triangle numbers 0 1 3 6 10 etc. and start on 1, 1 0 -2 -5 -9 is what you get. So that part is resolved.

[@fibroidss1194]

I think there’s something that does not really make sense what branch of the logarithm have you chosen? In the complex world, the exponential function is periodic on the imaginary axis it’s not invertible in C, you should consider a specific branch of the function before you can invert it

[@giuseppemalaguti435]

x=log(5)-25=iπ/ln5+2

[@rakenzarnsworld2]

No solutions

[@alextang4688]

It is obviously that x is an imaginary number, so we can use the polar form to solve it. 😉😉😉😉😉😉

[@diwakar500]

😮😮😢😅😅😮😮😮

[@adogonasidecar1262]

Complex, not imaginary

[@mcwulf25]

He posted it on the right channel then.

[@angeluomo]

This result shows that: x^((pi*i)/lnx)= -1 = e^(pi*i)