If you align the left and right expressions by aligning the shape of the binomial product by 2,3 and pay attention to the exponent, you can find that the exponent of the left side 2 is x = 1. Now x = 1 is It is certain that it is an answer. Even so, if you check if the index of 3 is 1, it will be an answer.
Good one! Hey, off-topic, but wanted to ask: I've been playing with cubics where the a term is 1 and the c term is 0, so x^3 + bx^2 + d = 0. I "discovered" (I'm sure I am not the first) a parameterization where if the roots are ak, a(k^2 - k) and a(1-k), the resulting cubic is x^3 - a(k^2 - k + 1)x^2 + (a^3)(k^2)((k - 1)^2) = 0. So if a and k are 2, we get x^3 - 14x^2 + 288 = 0, and the roots are 6, 12 and -4. Have you seen something like this? Is this going to generate all these depressed cubics, or just some? Is this useful to leverage to solve these? Anyway, I found it interesting and fun.
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If you align the left and right expressions by aligning the shape of the binomial product by 2,3 and pay attention to the exponent, you can find that the exponent of the left side 2 is x = 1.
Now x = 1 is
It is certain that it is an answer. Even so, if you check if the index of 3 is 1, it will be an answer.
Nice!
Second solution is approximately -5.17 😊
It's frighteningly close to -206797/40000.
@@QermaqLol I thought it was irrational number
@Qermaq wow that's more accurate.
@@rakenzarnsworld2 It is. It's that fraction plus something like 0,00005 something.
Good one!
Hey, off-topic, but wanted to ask: I've been playing with cubics where the a term is 1 and the c term is 0, so x^3 + bx^2 + d = 0. I "discovered" (I'm sure I am not the first) a parameterization where if the roots are ak, a(k^2 - k) and a(1-k), the resulting cubic is x^3 - a(k^2 - k + 1)x^2 + (a^3)(k^2)((k - 1)^2) = 0. So if a and k are 2, we get x^3 - 14x^2 + 288 = 0, and the roots are 6, 12 and -4. Have you seen something like this? Is this going to generate all these depressed cubics, or just some? Is this useful to leverage to solve these? Anyway, I found it interesting and fun.
X=1 by looking at it
Ah yes any engineer's favourite method: solution by observation
You can't solve by looking at it bro 😂
X= 1
You missed one
Just by sight, x = 1
x ≠ -2;
2^(x - 1) * 3^(3x/(x + 2) - 1) = 1
take log_2 of both sides:
(x - 1) + (2x - 2) * log_2(3)/(x + 2) = 0
multiply both sides by (x + 2) ≠ 0
(x - 1) (x + 2 + 2log_2(3)) = 0
x = 1 or
x = - 2 - 2log_2(3) = - log_2(36)
x = 1
x=1,x=-ln36/ln3...no,devo rifare i calcoli..x=-ln36/ln2=-5,17..ecco,this Is correct