An Interesting Exponential Equation

Рет қаралды 2,387

13 күн бұрын

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Пікірлер: 9

@fahrenheit2101 11 күн бұрын

Ooh, looks nice. As usual, here's my guess prior to watching. Given the powers of 3 present, I figured it would be a good idea to write x = 3^k, noting that x must be positive (if real) to ensure the RHS is positive. From this, we see that 3^27k = (3^3)^(3^2k) = (3^3)^(9k) so 3^2k = 9k, so 9^k = 9k, which has 2 solutions, k=1, and some smaller one. I think perhaps k = -W(-ln(9)/9)/ln(9) with W being the Lambert W function, but I might have messed that up, and that possibly provides the second solution, using another branch?. The actual values of x are therefore 3, and some smaller root, still greater than 1.

@mcwulf25 11 күн бұрын

I knew there would be a W. Left it to the end!

@scottleung9587 11 күн бұрын

Nice!

@phill3986 11 күн бұрын

😊😊😊 👍👍👍

@2012tulio 11 күн бұрын

X= {3 ; 1.188}

@rakenzarnsworld2 11 күн бұрын

x = 3

@DonEnsley-yi2ql 11 күн бұрын

Donain x 0 since base on left is x. Let a = 27 x^a = a^(x²) ln both sides. a ln x = x² ln a Divide by x² a. x⁻² ln x = ln a /a ln x e⁻² ˡⁿ ˣ = ln a / a Multiply by -2. -2 ln x e⁻² ˡⁿ ˣ = -2 ln a / a -2 ln x = W[(⅑) ln (⅑)] x = e ⁻⁽¹ᐟ²⁾ ᵂ⁽ ¹ᐟ⁹ ˡⁿ⁽¹ᐟ⁹⁾ ⁾ This is approximately 1.1880158846589424081914019779477681105487706109079210758564778249