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16

This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals. Assume $a,b\in\mathbb{R}$. Note that $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln ... 7 Taking the logarithm to base b of both sides, the equation is equivalent to$$(\log_bc)(\log_ba)=(\log_ba)(\log_bc)\ .$$I would say n^{\log_23} is better because it makes it clear that it is n to a constant power. This is of course also true for 3^{\log_2n}, but it's not so obvious. 5 HINT: First of all, notice that 15^{x-1} \equiv \frac{1}{15} \times 15^x, and hence:$$3^{2x} - 34\left(15^{x-1}\right)+5^{2x} \equiv 3^{2x} - \tfrac{34}{15}\left(15^x\right) + 5^{2x}$$Second, notice that$$u^2 - \tfrac{34}{15}uv + v^2 \equiv \tfrac{1}{15}(5u-3v)(3u-5v)$$If you put u=3^x and v=5^x, then notice that u^2 = 3^{2x}, uv = ... 3 They mean that$$y(x) = f_1(x) \times f_2(x) \times \ldots \times f_n(x)$$where none of the terms on the right-hand side are zero. Then you have$$ \ln |y| = \ln \left| \prod_{k=1}^n f_k(x) \right| = \ln \left( \prod_{k=1}^n \left| f_k(x) \right| \right) = \sum_{k=1}^n \ln \left| f_k(x) \right| $$Rewriting equivalently in more usual ... 3 Let A = 2^x. Then the equation becomes$$A^2 + 3A - 18 = 0$$which is just one of your everyday quadratic equations. Now, solve the quadratic equation for A to find that A = -6, 3. Equate both with 2^t to find that$$2^t = -6, 3$$For the case when 2^t = 3, we have 2^t = 3 \implies t = \log_2{3}. We reject the case for which 2^t = -6, ... 2$$\log{\binom n{\gamma n}} = \log{\frac{n!}{(n-\gamma n)!(\gamma n)!}}\\ = \left(n+\frac 12\right)\log n - n - \left(n\gamma +\frac 12\right)\log \gamma n + \gamma n \\- \left(n - n\gamma +\frac 12\right)\log (n - \gamma n) + (n - \gamma n) + O(1) \\= n\log n - (n - n\gamma)\log (n - \gamma n) - (n\gamma )\log \gamma n + \frac 12\log n - \frac 12(\log n + ...

1

The Taylor series of this iteration starts as \begin{align} f_1(x)&=x -\tfrac{1}{2}\,x^2 +\tfrac{1}{3}\,x^3 -\tfrac{1}{4}\,x^4 +\tfrac{1}{5}\,x^5 -\tfrac{1}{6}\,x^6 +\tfrac{1}{7}\,x^7 -\tfrac{1}{8}\,x^8 +\tfrac{1}{9}\,x^9 \\ f_2(x)&=x -x^{2} +\tfrac{7}{6}\,x^{3} -\tfrac{35}{24}\,x^{4} +\tfrac{19}{10}\,x^{5} ...

1

Perhaps you're trying to get at the Riemann sums resulting from subdividing each interval $[k, k+1]$ into $m$ equal-length pieces? This gives $$\frac{1}{k+1} = \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k+1} < \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k + \frac{i}{m}} < \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k} = \frac{1}{k},$$ so that $$... 1 A function f(x) is said to be non-zero if for any x \in \mathbb R f(x) \ne 0 ,i.e does not have a value 0. This is necessary because the function \log_a(x) is undefined at x=0. After that it is simply an application of the law,$$\log_k(ab) = \log_k(a) +\log_k(b)$$For proof,$$\text{Let }k^x = a, k^y = b\implies \log_k(a)=x, ...

1

I suspect the OP is using base 10 logs instead of the standard base $e$ log. So, using base 10 logs, We have that $\log(500) = 2.69897...$ and $\log(100)=2$. Thus, \begin{align*} &\quad 2 \log(x) + \log(5) = \log(500) \\ &\implies 2\log(x) = \log(500) - \log(5) \\ &\implies 2\log(x)=\log(100) \\ &\implies 2 \log(x) = 2 \\ &\implies ...

1

I do not know how to answer your question. However, in order you to challenge your challenger, I give you a few amazing results for $$f(n)=\int_0^\infty\frac{\log\left(\frac{1+x^a}{1+x^b}\right)}{\left(1+x^2\right)\log x}dx$$ in which $a=2n+\sqrt{4 n^2-1}$ and $b=n+\sqrt{n^2-1}$. $$f(1)=\frac{1}{4} \left(1+\sqrt{3}\right) \pi$$ $$f(2)=\frac{1}{4} ... 1 Firstly you should know that this is a direct result of the Logarithms of Powers Rule:$$\log(\sqrt{x}) = \log(x^{1/2}) = \frac{1}{2} \log(x)$$Here's a proof nonetheless, To Prove: \large \frac{1}{2} \log(x) = \log(\sqrt{x}) Proof: Let us assume that what we have is true. Let  \large \frac{1}{2} \log(x) = \log(\sqrt{x}) = k and let b be the base of ... 1 I'm supposeing that the square brackets are the INT function. The second equation supposes that all values are the same size. The correct form of it would be along the lines of the following. For eg 153, the first adds 7 to the total, while the second equation adds just 1. The equation as written below is a restatement of the equation above. ... 1 For all j such that 2^k\leq j<2^{k+1} for some integer k, we have [\log j]=k. Further note that there are exactly 2^k js that [\log[j]=k holds. Hence from$$0+1×2+2×2^2+3×2^3+4×2^4+5×2^5+6×2^6+7×2^7=1538, we can easily obtain ...

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