How would you solve the equation x^(ax+b) = c?

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How would you solve the equation x^(ax+b) = c?

can it be done? what's the steps?

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logramethics.

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The LHS is strictly increasing so just use the Bisection merhod.

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there is no hope, kill yourself

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>>7757666

A hint: Lambert W function.

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solve [math]ax+b = d \Leftrightarrow x = \frac{d-b}{a}[/math] and then try values for d. That's what calculators do really.

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It's simply unpossibles.

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>>7757719

/thread

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>>7757740

It's not. Consider the function f with f(x) = x^(2x+5). It's a coninuous function, f(1) = 1 and f(2) = 512 so by the intermediate value theorem where is an x between 1 and 2 with f(x) = 3.

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>>7757666

>> fzero(@(x)x.^(2*x+5)-3,1)

ans =

1.161843634045389

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[math]x = 3^{1/(2x+5)}[/math]

[math]x_0 = {-\infty, -1, 0, 1, \infty}[/math] pick whatever

iterate

[math]x_{n+1} = 3^{1/(2x_n+5)}[/math]

it converges quite fast

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>>7757666

underage

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can the W function be used here?

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>>7759649

You're right to use logarithms, but you should check your work when you're done.

1^(2*1+5) = 1^(7) = 1

1 != 3

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