How to solve x*a^x=b with math reasoning?

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The equation x*a^x=b cannot be solved exactly in general, so numerical methods like Newton's method are recommended for approximating roots. Start by selecting an initial guess, x0, and use the iterative formula x_{n + 1} = x_n - f(x_n)/f'(x_n) to converge to a solution. The discussion emphasizes the importance of understanding the behavior of the function by taking its derivative and analyzing the signs based on the values of a and b. For positive constants a and b, the approach involves checking the conditions for a and using a sign table to determine the existence of roots. Ultimately, knowing specific values for a and b simplifies finding the roots of the equation.
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x*a^x=b

how to solve this?
 
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You can't, in general, solve it exactly. Use a numerical method to approximate the roots.
 
This is a good candidate for Newton's method.
You first choose an x0 arbitrarily. You can graph it first, then choose an x0 wisely so it's near the roots.
Then use the formula:
x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}, and let n increase without bound to obtain the solution, i.e:
x = \lim_{n \rightarrow \infty} x_n.
Can you get this? :)
 
Yes.
And it works! :)
I didn't know about Newton's method before.
 
with math reasoning

Suppose a and b are unknown constants
Let y=xa^x and hence y=b
Take a ln of bothe sides leading to lny=lnx+xlna
We understand that x,a,b must be > 0

Taking a derivative of y gives us
\frac{dy}{dx}=(\frac{1}{x}+lna)xa^x
Now we find that x=0 (omitted), \frac{-1}{lna}

Next, we draw a table to check signs of \frac{dy}{dx}, but before that we check a
1. if 0<a<1
Look at the table and mark for sign (+/-), then check for y to compare with y=b (a.straight.line), which means you need to reason the value of b for where the root(s) exist.
2. if a>1
Do the same to find out root domain

Now things become easier when you know concrete constant a, b. just put them inthere to find a root. This way looks crary though :biggrin: but sovable domain can be understood
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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