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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?
>>
logramethics.
>>
The LHS is strictly increasing so just use the Bisection merhod.
>>
there is no hope, kill yourself
>>
>>7757666
A hint: Lambert W function.
>>
solve [math]ax+b = d \Leftrightarrow x = \frac{d-b}{a}[/math] and then try values for d. That's what calculators do really.
>>
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.
>>
>>7757666
>> fzero(@(x)x.^(2*x+5)-3,1)

ans =

1.161843634045389
>>
[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
>>
can the W function be used here?
>>
>>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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