Tagged Questions
2
votes
1answer
194 views
Why use radical notation instead of rational exponents?
I'm helping my younger sister for her math class. She has recently been taught integer exponents, and has starteed studying radicals (mainly square roots). The next topic will be rational exponents, ...
4
votes
3answers
324 views
How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution?
Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1 $$
non-numerically?
How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists ...
5
votes
2answers
140 views
How many positive roots does the equation $a^x=x^a$ have?
Let $a\in (1,e)\cup(e,\infty).$ I'd like to show that the equation $a^x=x^a$ has exactly two positive solutions, and one is larger and one smaller than $e.$ Is it even possible to show? I think I've ...
6
votes
3answers
474 views
What we can say about $-\sqrt{2}^{-\sqrt{2}^{-\sqrt{2}^\ldots}}$?
Problem:
How we can strictly prove $-\sqrt{2}^{-\sqrt{2}^{-\sqrt{2}^\ldots}}$ can't be 2?
Can $-\sqrt{2}^{-\sqrt{2}^{-\sqrt{2}^\ldots}}$ have the value expressed by complex numbers? (See below, in ...
39
votes
4answers
2k views
Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct?
Problem:
Find $x$ in
$$\large x^{x^{x^{x^{ \cdot^{{\cdot}^{\cdot}} }}}}=2$$
Trick:
$x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$, so,
$x^{(x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}})}=x^2=2$, and,
...
0
votes
1answer
103 views
How do I find p for equations of the form $\sum \limits_i \frac{a_i}{b_i^p} = 1$
The problem I'm facing is solving the following equation for $p$ given the constants $a_i$ and $b_i$:
$$
\sum_i \frac{a_i}{b_i^p} = 1
$$
Is there a general technique that would allow me to find a ...
11
votes
3answers
168 views
How do we solve $a \le b^{r}-r$ for $r$?
Given two values $a$ and $b$, how should one go about solving the following inequality for $r$:
$$a \le b^r -r .$$
Applying $\log_b$ on both sides of the inequality doesn't help me much since that ...
