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- čas přidán 20. 08. 2024
- A great exponential equation. Solution.
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Should mention that x=log ф
Wow! I actually understand that. Miracles will never cease.
Answer x = log 1.61803 or 0.2089 or 0.21
This is the golden ratio
1^x + 10^ x =100^x
1 + 10^x = 100^x (since 1^ any number = 1)
0= 100^x - 10^x -1
let 10^x = p
Hence,
0= p^2 - p -1
Using the quadratic formulae and
p = (1 + or -sqrt 5 )/2 or 1.61803 or -0.61803 Golden ratio
Hence, 10^x = 1.61803 as you find the log a negative number, so disregard - 0.61803
x log 10 = log 1.61803 find the log of both sides
x = log 1.61803 or 0.2089865693 Answer
Y = 10^x ---> y^2 = y + 1
For the 2nd root, we can choose for the branch cut of the logarithm in the complex plane one that's not on the half-line of non-positive reals. E.g., choose the branch cut on the upper imaginary line such that the argument of a complex number, θ, satisfies -3π/2
Shouldn't we instead of assuming -1=e^(iπ) or -1=e^(iπ) assume -1=e^(inπ) with n being any arbitrary odd integer ?
@@Apollorion Yes, in solving y²-y-1=0 & then x, I think you can choose other branches of ln, as you suggest. A problem, though, is that the initial assumption 1^x=1 is not valid in general even for real x.
x.ln(1)+x.ln(10)=2.x.ln(10)
Si x non égal à 0 alors 1=2 donc "x=0" 😅
100/1 = 100100/10=10 {100+10}= 110 10^10^10 2^52^52^5 1^1^1^1^2^1 2^1 (x ➖ 2x+1)
let u=10^x , u^2-u-1=0 , u=(1+V5)/2 , 10^x=(1+V5)/2 , x=ln((1+V5)/2)/ln10 , test , 1+10^(ln((1+V5)/2)/ln10)=~ 2.61803 ,
100^(ln((1+V5)/2)/ln10)=2.61803 , same , OK ,
The initial assumption 1^x=1 isn't valid in general, even for real x. E.g., 1^(½)=±1 or 1^(⅓) has complex roots. In fact, the solution presented contradicts this assumption, b/c
x≅0.2 ⇒ 1^x≅e^(0.2·2niπ) for an integer n, which isn't equal to 1 if n≠0.
So, we haven't fully solved the initial equation, even for real x.
How about this?
I'm still confused with this topic. This (below) is the logic way that appeared in my mind, (but yeah, I'm still confused is it correct or not).
We know that
aⁿ ⎛ a ⎞ ⁿ
------- = | ------ |
bⁿ ⎝ b ⎠
*Then*
10ˣ ⎛ 10 ⎞ ˣ
-------- = | -------- | = (1)ˣ
10ˣ ⎝ 10 ⎠
10ˣ
But -------- is certainly equals to (1).
10ˣ
*So, (1ˣ = 1). We can substitute (1)ˣ with (1).*
But I agree with you, 1ˣ doesn't always equals to 1.
Perhaps we can say that 1ˣ can be substituted with 1 for some cases?, for example: the case what I've said above.
(a/a) of course = 1
Then, what if we change "a" with 10ˣ?
So (10ˣ/10ˣ) = ?
I don't know, but I still think it's 1, although we can write it as 1ˣ too.
*Btw, I'm so sorry for my bad English. I can't speak English well, but I hope you can understand what I mean.
@@TidakTerdefinisi 10^x, in general, isn't a single-valued function, & the notation is confusing this. Take the simpler example of 4^(½). It's double-valued. So when you right 4^(½)/4^(½), it's not clear if you're dividing the same branches of the square root or different ones. It could be 2/2 or (-2)/(-2) or 2/(-2) or (-2)/2. If you write √4/√4, then it's clear b/c that's the positive root.
@@TidakTerdefinisi In your notation, presumably, "a" is a fixed number; has a single value. But 10^x is not single-valued.
Had you not mentioned it, wouldn't have known you're not a native English speaker.
@@kyintegralson9656 Oh, I see, yes you're right, it makes sense. If "a" doesn't have a single value, the result of (a/a) can't be simplified as 1, because we don't know what we try to solve, is it (a₁/a₁) or (a₁/a₂) or etc?, and we can't easily assume that it must be (a₁/a₁) or (a₂/a₂) because we don't have basic evidence to prove it must be "the same a". Thank you so much, Brother.
Maybe, I can conclude it as: if "a" has more than one values, so _(a/a) = 1_ is just one of the solutions, (not the only one), because there are some others solutions.
Btw, I tried to ask WolframAlpha just for fun, (10ˣ/10ˣ), and that site gave me: the result is 1. Oh my, I hate (1ˣ), haha 😅
Btw, there is one more thing that make me confused. Do you mind if I ask it to you? That's about "i". Does _i = √(-1)_ (positive, only one value), or _i = ± √(-1)_ (have two values)? Because both of (√(-1))² and (- √(-1))² are equal to (-1). What about (i/i) ?
By inspection my guess was .2, but knew their were more trailing digits. Watched your solution and by slide rule .209
thanks for your lesson! The explanation is detailed and clear. Where can I see more of your lessons on other topics? Thank you in advance and good luck with everything.
And once again the golden ratio shows up 😉
as early as possible, set if we are working in Real or Complex
I solved this problem really 😎
without watching this video.
I only got halfway before watching it: 1 + 10^x - 10^2x = 0
I also
So whatdowehaverighthere😂…
Answer is ln(phi) / ln(10)
That's what I got.
(log_e((1+sqrt(5))/2))/log_e(10) = log_10(1+sqrt(5))-log_10(2) .
@@user-wl4zu2ok1e Yeah, using the Change of Base formula, the answer becomes log(Φ).
I reckon it's going to have to do with phi... Curious one
Yeah, I noticed that too.
For me the interesting thing is that this result isn't confined to base 10.
If you solved any problem "a^0 + a^x = a^2x" the answer would be log_a(\phi). Yet another magical property of phi I guess.
So boring!
1ˣ + 10ˣ = 100ˣ ........... *(x = ?)*
1ˣ + 10ˣ = (10²)ˣ
1ˣ + 10ˣ = (10ˣ)² ......... *Let: (y = 10ˣ)*
1ˣ + y = y²
*We know that*
aⁿ ⎛ a ⎞ ⁿ
------- = | ------ |
bⁿ ⎝ b ⎠
*Then*
10ˣ ⎛ 10 ⎞ ˣ
-------- = | -------- | = (1)ˣ
10ˣ ⎝ 10 ⎠
10ˣ
But -------- is certainly equals to (1).
10ˣ
*So, (1ˣ = 1). We can substitute (1)ˣ with (1), and we get*
*1* + y = y²
y² - y - 1 = 0 ......... *(factor)*
D = b² - 4ac = (-1)² - 4(1)(-1) = 1 + 4 = 5
(-b) ± (√D) 1 ± (√5)
y = ------------------ = ----------------
2(a) 2
*Since (y = 10ˣ), so*
y₁ = (10)ˣ
⎛ 1 + (√5) ⎞
x = log(y₁) = log | --------------- | = *log(1 + √5) - log(2)*
⎝ 2 ⎠
y₂ = (10)ˣ
⎛ 1 - (√5) ⎞
x = log(y₂) = log | --------------- | = *undefined*
⎝ 2 ⎠
because 1 - (√5) = negative, but negative number can't be located inside log, so it's not the answer.
*Well, the answer is: x = log(1 + √5) - log(2)*
Well perfect