Algebraic Solution for x + 3^x = 4

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The equation x + 3^x = 4 can be approached by defining f(x) = x + 3^x - 4 and verifying that f(1) = 0, indicating x = 1 is a solution. To confirm that this is the only solution, one can analyze the monotonicity of f(x). While the Lambert W function provides a method to express the solution, it is not purely algebraic. Ultimately, the algebraic solution for this equation is not feasible, reinforcing that x = 1 is the only solution.
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how can I solve this using algebraic methods? I know the solution is x = 1 from just looking at it, but not sure how to do it algebraically:

## x + 3^x = 4 ##
 
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You can set f(x) = x+3^x-4, take x=1 as a solution to f(x) = 0 and look if it is monotonic to show that it has no other solutions. Algebraically is not possible.
 
zoxee said:
how can I solve this using algebraic methods? I know the solution is x = 1 from just looking at it, but not sure how to do it algebraically:
## x + 3^x = 4 ##
a very strange way of writing one:

x=\frac{-W(3^4\ln(3))}{\ln(3)}+4=1
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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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