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I tried many ways of solving this, eventually getting to $\frac{3x+1}{x+4}=\frac{\ln 2}{\ln 3}$. This didn't lead anywhere. This is fine. Did you solve this expression for $x$? $$\frac{3x+1}{x+4}=\frac{\ln 2}{\ln 3} \iff x = \frac{\ln 3 -4\ln 2}{\ln 2 - 3\ln 3} = \ldots$$ Now use properties of logarithms to rewrite this expression towards the desired form....

2

Your last line is ok, then multiplying out we get $$3x\ln(3)+\ln(3)=x\ln(2)+4\ln(2)$$ and now combining like terms: $$x(3\ln(3)-\ln(2))=4\ln(2)-\ln(3)$$ Can you finish?

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You are right: Lognormal distribution is not representative of stock prices. Neither is it verified by backward-looking evidence as shown in your graphs, nor it is supported by forward-looking market information, such as the stock-option prices. It is well known that stock prices exhibit the so-called fat tails, again verifiable from historical data as ...

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You are differentiating w.r.t. $x$ but you are supposed to differentiate w.r.t. $s$. The dertivative of $x^{s}$ w.r.t. $s$ is $x^{s} \ln x$.

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Note that $\frac{n!}{a_1!a_2!\ldots a_k!}$ counts the number of functions $f:\{1,\ldots,n\}\rightarrow \{1,\ldots,k\}$ such that there are $a_i$ elements mapping to $i$ - it can be understood as the multinomial coefficient ${n \choose a_1,a_2,\ldots,a_k}$. Then, if we fix some $k$ and sum this quantity up over all possible decompositions $a_1+\ldots+a_k=n$, ...

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Hint: $$\frac{147}{36}=\frac{49}{4\cdot 3}=\frac{7^2}{2^2\cdot 3}.$$

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$$\log \frac{147}{36}$$ $$=\log\frac{3\times7^2}{2^2\times 3^2}$$ $$=\log\frac{7^2}{2^2\times 3}$$ $$=2\log 7-2\log 2-\log 3$$ $$=2c-2a-b$$ Hope it helps:)

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OK, I'm not super happy with what I have, but I'm posting in case it helps others find a better formula. $$\boxed{I(x)=\pi\log 2 -\frac 1 2 \int_x^1 \frac{K(\sqrt{1-t})-\frac \pi 2}{1-t}dt=\pi\log 2-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1}}$$ where we identify the complete elliptic integral of the first ...

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This isn't right though, OP. Given $y=p(x)$ where $p$ is a polynomial, the asymptotic slope of the log-log plot is the degree of $p$; the lower-order terms vanish in their effect on the log-log plot for $x$ large. The log-log plot of $y=8x^3$ would have a slope of 3; e.g., $\log y =3 \log x + \log 8$. The log-log plot of $y=8x^3+7x^2+6x+5$ would be a curve ...

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A double-log coordinate system makes power functions: $$y = ax^b$$ become straight lines. The parameter $b$ determines the visual slope on the line; $a$ determines its location (up/down). Taking the logarithm on both sides of $y=ax^b$ gives you $$\log y = \log a + b\log x.$$ You can now set $y'=\log y$ and $x'=\log x$ for physical plotting distances and ...

1

Proof 1: Show the equality from the right to the left. It's simply $$\mathrm{e}^{\ln(a^x)}=\mathrm{e}^{x\ln(a)}=\mathrm{e}^{\ln(a)x}=(\mathrm{e}^{\ln(a)})^x=a^x$$ provided $a>0$ and $x\in\mathbb{R}$. Proof 2: Let $a,b>0$ and $a\neq 1$ and consider $a^x=b$. Then you can define $x=\log_a(b)$, in case of $a=\mathrm{e}$ you can abbreviate $\ln(b)=\log_{\... 5 By definition,$\ln{a^x}$equals a number that when$e$is raised to that number, you get$a^x$. So, let's say that$\ln{a^x}=y$. Then,$e^y=a^x$. But we know what$y$is in that last expression. It's equal to$\ln{a^x}$. Thus,$e^{\ln{a^x}}=a^x$. 6 The function$\ln:(0,\infty)\to\mathbb R$is by definition the inverse of exponential function$\mathbb R\to(0,\infty)$. So: $$y=e^{\ln(y)}\text{ for every }y\in(0,\infty)$$ Now substitute$y=a^x$. 2 Depending on how you define$\ln x$, one possible way is to define$f^{-1} (x) = \ln x$as the inverse function of$f(x) = e^x$. Then: $$e^{\ln(a^x)} = f\Big(f^{-1} \left(a^x \right) \Big) = a^x$$. 1 Remember that$\ln{e}=1$and$a\ln{x}=\ln{x^a}$. Therefore: $$2=2\cdot 1=2\cdot\ln{e}=\ln{e^2}.$$ So, what we have now is: $$\ln{\frac{x-2}{x-3}}=\ln{e^2}.$$ It's easy to see that the two sides of the equation are gong to be equal to each other only if the expressions inside the logarithms are equal. Thus, the previous equation is equivalent to this ... 2 It is not holding well. In fact,$\ln\left(\frac{x-2}{x-3}\right)=2\iff\frac{x-2}{x-3}=e^2$. And$\ln\left(\frac{x-2}{x-3}\right)\neq\ln_{x-3}(x-2). 1 Well, we know that: $$\log_\alpha\left(\beta\right)=\frac{\ln\left(\beta\right)}{\ln\left(\alpha\right)}\tag1$$ So, in your example we are trying to solve: $$\log_\alpha\left(x+1\right)+\log_\alpha\left(x-1\right)=\text{n}\tag2$$ Now, we rewrite: $$\frac{\ln\left(x+1\right)}{\ln\left(10\right)}+\frac{\ln\left(x-1\right)}{\ln\left(10\right)}=\frac{\ln\... 6 Careful: it is \log(x^2-1) and not \log(x-1)^2. And you forgot that from x^2 =a you got x=\pm \sqrt{a} Finally, you should check if your solution is >1 (because of domain of \log(x-1)) 2 I tried to think of how a middle schooler might solve this: Let x=2^a. Then we have$$\log_3 (2^a+1) + a = 5.$$So that$$2^a+1 = 3^{5-a}.$$Multiply by 3^a to get$$6^a+3^a = 3^5 = 243.$$If the student knows that 6^3 = 216, he knows that a is pretty close to 3, which does in fact work, giving x=8. 4$$\log_3(x+1)+\log_2 x=5\log_3(x+1)=5-\log_2 xx+1=3^{5-\log_2 x}$$For integer solutions for x we must have:$$5-\log_2 x>0$$⇒$$\log_2 x<5Therefore we must check numbers 4, 3,2, 1 which gives: \log_2 x= 1, 2, 3, 4 ⇒x=2, 4, 8, 16 These solution must also satisfy the initial equation; corresponding values are: \log_3 (x+1)=5-... 2 Let f(x)=\log_3(x+1)+\log_2x. Thus, since f increases, our equation has one root maximum. 8 is a root, which says that it's an unique root and we are done. 0 I don't know if this is what you want, but I hope it helps. log_2(x)=a means that 2^a=x log_3(x+1)=b means that 3^b=x+1 To get a+b=5 where 3^b is after 2^a by +1 only (somehow consecutive numbers), then it is most likely that a and b are integers. (This step isn't reasonable actually, it is just "looks like so") Now if a and b ... 0 \begin{align} I&=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2(1-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2x}{1-x}\ dx\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\ &=2\left(3\zeta(2)\zeta(5)-\frac92\zeta(5)\right)-2\zeta(5)\\ &... 0 \require{begingroup} \begingroup \def\W{\operatorname{W}}\def\e{\mathrm{e}} \begin{align} \frac{a\cdot (b+x)}{c} = (1+\frac{a\cdot x}{c}) \cdot \ln(1+\frac{a\cdot x}{c}) \tag{1}\label{1} \end{align} Indeed, the solution to equation \eqref{1} can be expressed in terms of the Lambert W function. The usual approach in this case is to transform ... 0 More generally for positive real a,b,d\ln{\big|da+db\big|}=\ln{\big|d(a+b)\big|}=\ln{\big|d\big|}+\ln{\big|a+b\big|}=c_1+\ln{\big|a+b\big|}$$where a=x,b=2,d=2 in your example. Then, if e=-\frac{3}{2} and f=\frac{1}{2},$$fa+\ln{\big|da+db\big|}+c_2=fa+\ln{\big|a+b\big|}+c_1+c_2=fa+\ln{\big|a+b\big|}+c_3$$so the difference is only by a ... 1 \ln|2x+4|=\ln|2(x+2)|=\ln|2|+\ln|x+2|=\ln|x+2|+C 5 Both answers are the same, mind the constant:$$\ln|2x+4|+\color{blue}{C_1}=\ln|2\left(x+2\right)|+\color{blue}{C_1}=\ln|x+2|+ \underbrace{\ln 2 + \color{blue}{C_1}}_{\color{purple}{C_2}} = \ln|x+2|+\color{purple}{C_2}$$1$$\begin{align}\frac12 x + \frac 32 \ln |2x + 4| + C_1 &= \frac12x + \frac32\ln(2\cdot|x+2|) + C_1\\&=\frac12x + \frac32(\ln 2 + \ln|x+2|)+C_1\\&=\frac12x + \frac32 \ln|x+2| + (\frac 32 \ln 2 + C_1)\\&=\frac12x + \frac32\ln|x+2| + C_2\end{align}$$1 The whole idea of math is to simplify something so that it becomes usable. Knowing that the derivative is constant(1) is way more useful then having this expression which might not be constant, that you'd have to evaluate for each x. In general my advice is don't skip anything in math because chances are it will either appear in future problems or it ... 2 In the paper Zujin Zhang Mean-value property of the heat equation the related calculations are carried out in detail. 1 When you can, it's usually a good idea to simplify. Think of it as becoming a habit because often, expressions such as \sqrt[3]{x^3} might only arise as intermediate steps and for the rest of the calculations or work to come, you make life easier by simplifying when and where you can - as a general rule of thumb. However, there is something I don't see ... 2 \log_{2}3 = a means that 2^a = 3. By cubing both sides we get (2^a)^3 = 3^3 \Rightarrow (2^3)^a = 3^3 \Rightarrow 8^a = 27 \Rightarrow \log_{8}27 = a 0 Because$$\sum_{cyc}\log\frac{x^2}{y^2}=\sum_{cyc}(\log x^2-\log y^2)=\sum_{cyc}(\log x^2-\log x^2)=0.$$0$$y = a_0\exp\left(-\frac x{a_1}\right) + (1-a_0)\exp\left(-\frac x{a_2}\right) + a_3\tag{1}$$An easy method (no iterative calculus, no initial guess) is explained in the paper https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales The case of functions made of a sum of exponentials is treated pages 71-73 with a numerical example in case ... 3 \log a+\log b =\log (ab) for all a ,b >0. So LHS is \log 1. 0$$\frac{\log x^x}{xy-xz}=\frac{\log y^y}{zy-xy}=\frac{\log z^z}{xz-yz}=k$$So,$$\log(x^x)=k(xy-yz)\log(y^y)=k(yz-yx)\log(z^z)=k(xz-yz)$$Add all these and get the result. Hope it helps:) 2 As I mentioned in my comment, you have enough information in the edits you've made to answer this. Specifically, for integers a,b \geq 2, the assumption that \log_ab is rational implies there are positive integers p,q such that a^p = b^q, so the prime factorizations of a and b must have the same primes, e.g.$$ a = p_1^{k_1} p_2^{k_2}\cdots p_n^{... 0 is it\log \frac{n}{n+1} $? if yes then the first method you did something wrong$\log \left(\frac{n}{n+1} \right) = \log (n) - \log (n+1)$you cant simplify further$\log(n+1) \ne \log(n) \log (1) $and in the second method you used the divergence test which just tells you if the limit is not zero then the series is diverge but if it is zero you need to ... 5 For$\Re(s)> 0$, let$F(s)$be defined as$$\sum_{n=0}^\infty (-1)^n\left[(n+1)^{1-s} + \frac 1 2 (n+1)^{-s} + (n+1)(n+2)\left[(n+1)^{-s}\ln(n+1) - (n+2)^{-s}\ln(n+2)\right]\right].$$ Then$F(s)$is an analytic function and$\displaystyle S= \lim_{\substack{s\to 0\\\Re(s)>0}}F(s)$holds. Assume$\Re(s)>3$so that the sum converges absolutely. Then ... 2 Splitting up the positive and negative terms gives $$S=\sum_{n=1}^\infty-1+2n(2n-1)\ln\left(1-\frac1{2n}\right)-2n(2n+1)\ln\left(1-\frac1{2n+1}\right)$$ Since$\displaystyle\ln\left(1-\frac1n\right)=\ln(n-1)-\ln(n)=-\int_0^1\frac1{x+n-1}~\mathrm dx$we provide the more suitable form $$S=\int_0^1\sum_{n=1}^\infty-1+\frac{2n(2n+1)}{x+2n}-\frac{2n(2n-1)}{x+... 0 Using AM >= GM (a+b)/2 >=sqrt (ab) log(base a) of (( a+b)/2) >= 1/2(1 + log(base a) ( of b)) log(base b) of ((a+b)/2) >= 1/2(1+ log (base b ) (of a )) Adding both LHS >= 1 +(1/2) ( log (base a) (of b) + log ( base b )( of a)) Again using AM>> GM on log terms on RHS we will get LHS >= 2 0 By C-S and AM-GM$$\log_a\frac{a+b}{2}+\log_b\frac{a+b}{2}=\frac{1}{\ln_{\frac{a+b}{2}}a}+\frac{1}{\ln_{\frac{a+b}{2}}b}\geq\geq\frac{(1+1)^2}{\log_{\frac{a+b}{2}}a+\log_{\frac{a+b}{2}}b}=\frac{4}{\log_{\frac{a+b}{2}}ab}\geq\frac{4}{\log_{\frac{a+b}{2}}\left(\frac{a+b}{2}\right)^2}=2.$$1 By AM-GM we have$$\frac{a+b}2\geq \sqrt{ab} $$and since \log is an increasing function for bases >1 we have$$\log_a(\frac{a+b}2) +\log_b(\frac{a+b}2) \geq \log_a(\sqrt{ab}) +\log_b(\sqrt{ab}) =1/2(2+\log_a(b)+\log_b(a))$$And we have that \log_a(b) +\log_b(a) =\log_a(b) +\frac1{\log_a(b)} \geq 2 because x+\frac1 x \geq 2 2 Let's set f(x)=\dfrac 1{\ln(x)} We have f''(x)=\dfrac{2+\ln(x)}{x^2\ln(x)^3}>0 on (1,+\infty), thus f is convex. The result is just an application of the convexity inequality$$f\left(\frac{a+b}2\right)\le \frac 12f(a)+\frac 12f(b)\iff \frac 1{\ln(a)}+\frac 1{\ln(b)}\ge \frac 2{\ln(\frac{a+b}2)}$$3 We have to prove that$$\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(a)}+\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(b)}\geq 2$$this is$$\ln\left(\frac{a+b}{2}\right)(\ln(a)+\ln(b))\geq 2\ln(a)\ln(b)$$or$$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq 2\ln(a)\ln(b)$$Now we have by AM-GM:$$\frac{a+b}{2}\geq \sqrt{ab}$$taking the logarithm on both sides$$\ln\left(\... 3 Using the change of base formula the inequality is equivalent to $$\frac{1}{\ln a}+\frac{1}{\ln b}\ge \frac{2}{\ln\big(\frac{a+b}{2}\big)}.$$ This, in turn, is equivalent to the function $$f(x)=\frac{1}{\ln (c-x)}+\frac{1}{\ln x}$$ being minimized on$(1,c-1)$at$x=c/2$. As$f$is unbounded (towards$+\infty$) on this interval, if it has only one ... 1 Assuming you mean in Newton's notation $$x^2y''=2y,$$ this is a Cauchy-Euler equation where you can try to find basis solutions in the form$y=x^m$, which gives a characteristic equation in$m$, $$0=m^2-m-2=(m-2)(m+1)$$ so that the solution basis is indeed$x^{-1}, x^2$. You get an additional constant because you construct$z$using the first ... 1 Another Hint A differential equation is any equation containing one or more derivatives. $$x^2y_2=2y$$ isn't a differential equation as it doesn't contain any derivatives. The only thing that you could solve for$x^2y_2=2y$would be finding$x,y,y_2$in terms of the other two variables. 1 HINT If you plug your solution in the original DE, you will see it is not true for all values of$E$. 0 Calculate$\sqrt [ 4 ]{ 136 }$using interpololation.$3^4 = 814^4 = 256256 - 81 = 175136-81 = 55\sqrt [ 4 ]{ 136 } \approx 3 + \frac{55}{175}$We started with rough estimations, but now we're going to tenths:$34^4 = 133633635^4 = 15006251500625 - 1336336 = 1642891360000 - 1336336 = 23664\sqrt [ 4 ]{ 136 } \approx 3.4 + \frac{...

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