I Solved An Interesting Exponential Equation

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[mcwulf]

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]

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.

[allanmarder]

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

[foxyone]

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

[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))

(i((π+2πk))/(ln(5)))+2
k its an integer

[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?

[levskomorovsky]

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

[fibroidss]

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

[alextang]

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

[angeluomo]

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