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5

While I don't believe there is a nice closed form for $S_n$, you can write the sum in terms of known functions and constants up to a very small error. Specifically, $$\sum_{i=2}^{N}\log_{i}(N)=\text{li(N)}\log N+C\log N+O(1),$$ where $\text{li}(N)$ is the logarithmic integral and $C$ is a constant equal to ...


3

If your inequation is $$ \log_2\left(1+\frac{1}{x-1}\right)<1 $$ then it is equivalent to the system of inequations \begin{cases} 1+\dfrac{1}{x-1}<2\\[2ex] 1+\dfrac{1}{x-1}>0 \end{cases} because $2>1$ and so the exponential with base $2$ is increasing. The first is obtained by removing the logarithm, the second because you need to ensure the ...


2

Suppose $$\log_{a} (b) = x.$$ Then, by definition, we have $$b = a^x,$$ where $a>0$. Now let $c>0$. Taking the logarithm to the base $c$ of both sides of the equation $b = a^x$, we get $$\log_{c} (b) = \log_{c} (a^x).$$ Or $$\log_{c} (b) = x \log_{c} (a)$$ using the property of the logarithm. So, if $\log_{c} (a) \ne 0$, then upon dividing both ...


1

By using the identity $$ \log_b(a)=\frac{\ln{a}}{\ln{b}} $$ Your equation becomes: $$ \frac{\ln (x y)}{\ln (x y z)}+\frac{\ln (x z)}{\ln (x y z)}+\frac{\ln (y z)}{\ln (x y z)}$$ Which you could further simplify to (using the identity $\ln{(a\times b)}=\ln{a}+\ln{b}$: $$ \frac{\ln (x y)+\ln (x z)+\ln (y z)}{\ln (x y z)}= \frac{2\ln (x)+2\ln (y)+2\ln (z)}{\ln ...


1

The formula used for the spiral is $r=a\cdot e^{b\cdot\phi}$. The arc below and above a selected arc have radii $r_\pm=a\cdot e^{b\cdot(\phi\pm2\pi)}$. The radius of a circle on the arc should be proportional to $r$, say $c⋅r$, with the condition that circles on neighboring arcs do not overlap, $$ r_-+c⋅r_-\le r-c⋅r\land r+c⋅r\le r_+-c⋅r_+ $$ where both of ...


1

If you can find the prime factorization of the number, take the greatest common divisor of all the exponents in it. Unfortunately factoring large numbers is not quick, so simply checking all possible degrees up to $\log_2$ of the number might well be faster asymptotically. For most inputs, a combination might be the best strategy -- look for small prime ...


1

Equation $$f(x)=7^{(2x+1)} + (2(3)^x) - 56 = 0$$ cannot be solved using elemental functions. What you can observe is that $f(0)=-47$ and $f(1)=293$. So, there is a solution for $0<x<1$. If you further refine, you could notice that $f(1/2)=2 \sqrt{3}-7=-3.5359$, so the solution is pretty close to $x=0.5$ (just above since $f(x)$ varies extremely fast). ...


1

Notice $$ \ln x = 2 \ln y \iff \ln x = \ln y^2 \iff x = y^2 $$ and $$ 3^x = 27 y \iff 3^x = 3^3 y \iff 3^{x-3} = y \iff 3^{y^2 - 3} = y \iff \ln 3 ( y^2 - 3) = y \iff y^2 - 3 - \frac{y}{\ln 3} = 0 \iff y^2 - \frac{1}{\ln 3} y - 3 = 0$$ This is a quadratic equation, which you can easily solve.


1

A simple (perhaps useful, perhaps not) bound via Jensen inequality. Because $1/\log(x)$ is convex: $$S_n = \sum_{i = 2}^{n}\log_i{(n)}= (n-1) \log(n) \frac{\sum_{i = 2}^{n} \frac{1}{\log(i)}}{n-1} \ge (n-1) \log(n) \frac{1}{\log \frac{\sum i}{n-1}}= \frac{(n-1) \log(n)}{\log(\frac{n}{2}+1)} $$ This bound seems quite decent. For large $n$ it tends to $n$ ...


1

Isn't it logarithmic in the dimension since the $\log n$ is there in both cases? The presence of $\log n$ does not make something logarithmic in $n$. The bound of the form $)(\log n)$ does. Since it is not true that $n^{6/5}\log n =O(\log n)$, the expression is not logarithmic. It is not polylogarithmic either, since there is no bound of the form ...



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