Intereting Posts

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Does this characterize compactness?
Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$

This equation clearly cannot be solved using logarithms.

$$3 + x = 2 (1.01^x)$$

Now it can be solved using a graphing calculator or a computer and the answer is $x = -1.0202$ and $x=568.2993$.

- Can this function be rewritten to improve numerical stability?
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- General solution to expressions, without calculating exact roots (A generalization of Newton's identities)
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- Proving that $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$
- Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable

But is there any way to solve it algebraically/algorithmically?

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- Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum
- Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
- Can I disprove a claim using the contrapositive?
- Building the integers from scratch (and multiplying negative numbers)
- Finding how multiplication and addition behave on $\mathbb{F}_4$ without any result
- How to find $f$ and $g$ if $f\circ g$ and $g\circ f$ are given?
- How to solve $\sin x +\cos x = 1$?
- Factoring with fractional exponents

I have solved a question similar to this before. In general, you can have a solution of the equation

$$ a^x=bx+c $$

in terms of the Lambert W-function

$$ -\frac{1}{\ln(a)}W_k \left( -\frac{1}{b}\ln(a) {{\rm e}^{-{\frac {c\ln(a) }{b}}}} \right)-{\frac {c}{b}}

\,.$$

Substituting $ a=1.01 \,,b=\frac{1}{2}\,,c=\frac{3}{2}$ and considering the values $k=0$ and $k=-1$, we get the zeroes $$x_1= -1.020199952\,, x_2=568.2993002 \,. $$

Polynomials don’t play nice with exponentials, so no. If you work hard, you might find an answer in terms of the Lambert W function, but if I did I wouldn’t feel much more enlightened.

A standard root finding procedure (such as Newton’s method) should solve the problem for you. You might also be interested in the Lambert W function, which will give you a “closed form” solution, assuming you have access to that function of course.

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