# Tag Info

7

Hint: $\ln(n-1)-\ln(n)=\ln\frac{n-1}{n}$. What is $\lim_{n\to\infty} \frac{n-1}{n}$?

5

$$0\lt\ln(n)-\ln(n-1)=\int_{n-1}^n{dx\over x}\lt\int_{n-1}^n{dx\over n-1}={1\over n-1}$$ Now squeeze.

5

Hint. You may transform the equation $$\frac{x-1}{e^x-1} = y \tag1$$ with a little algebra into $$-(x+y-1)e^{-(x+y-1)}=-ye^{1-y},\tag2$$ set $X:=-(x+y-1)$ obtaining $$Xe^X=-ye^{1-y} \tag3$$ then use the Lambert function W to get $$x=1-y-W\left(-ye^{1-y}\right). \tag4$$

4

$$\dfrac{(2\ln a- \ln b - 5\ln c)}{2}$$ $$=\dfrac{(\ln a^2- \ln b - \ln c^5)}{2}$$ $$= \dfrac{\left(\ln \dfrac{a^2}{b} - \ln c^5\right)}{2}$$ $$= \dfrac{1}{2}\ln \dfrac{a^2}{bc^5}$$ $$= \ln \dfrac{a}{\sqrt{bc^5}}$$

3

$$x=\log_2(2^x)$$ $$f(x)=\log_2[2^x(2^{x+2}-5+2^{-x+2})]=\log_2[2^{2x+2}-5\cdot2^x+4]$$ Now $4\cdot2^{2x}-5\cdot2^x+4=(2^{x+1})^2-2\cdot2^{x+1}\cdot\dfrac54+\left(\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $=\left(2^{x+1}-\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $\ge4-\left(\dfrac54\right)^2$

3

Draw a graph. $\log x < 2x$ A proof is by noting that $\log x < 2x$ for $x < 1$ and then differentiating both sides to see that the LHS grows slower than the RHS. Equivalently, $e^{2x} > x$

2

$$\lim_{x\rightarrow \infty }\ln(1-1/n)=\ln(1)=0$$

2

Hint: $$\log a -\log b= \log \frac{a}{b}$$ Which leads to $$\lim_{n\to\infty} \log \frac{n+1}{n}=\lim \log\left(\frac{1+\frac 1n}{1}\right)=\log\left(\lim 1+\frac 1n\right)=\log 1=0$$

2

$$\lim_{r\rightarrow 0^+} r\ln\ r = \lim_{r\rightarrow 0^+} \frac{\ln\ r}{1/r}=\lim_{r\rightarrow 0^+} \frac{\frac{1}{r}}{-\frac{1}{r^2}} =0$$ Then $$|y\ln\ (x^2+y^2) - 0 | \leq |r \ln\ r^2|=2r|\ln\ r|\rightarrow 0$$ where $r:=\sqrt{x^2+y^2}\rightarrow 0$.

1

Considering $x=rcos(\theta)$ and $y=rsin(\theta)$ would lead you to lim_{r\to 0}$$2rsin(\theta)ln(r) that tends to zero as r tends to 0. 1 Consider the situation at time t where there are n_i(t) infected computers and n_c(t) clean computers. We know that n_c(t)+n_i(t)=N, the total number of computers. The infected computers send out two worms each. The first worms could infect n_i(t) computers. But some of the computers that receive worms are already infected. The number of clean ... 1 There are definitely tricks. I have no idea whether this solution is the intended, but anyway, here goes. Lets start with the last equation and tweak it a bit.$$\begin{align*} \log(zx)-\log z\log x &=0 \\ 1 - \log z -\log x +\log z\log x &= 1\\ (1- \log x)(1- \log z) &= 1.\end{align*}$$I would then let X = 1- \log x, Y = 1- \log y and Z = ... 1 Using Polar coordinates. Let x = r\cos(\theta)\ and \ y=r\sin(\theta) hence the function becomes \rightarrow \frac{\ln( 1 +r^3\cos^3(\theta)sin^3(\theta))}{r} . As r^3\cos^3(\theta)\sin^3(\theta) \rightarrow 0 as r \rightarrow 0 hence \ln(1 + r^3\cos^3(\theta)\sin^3(\theta)) \sim r^3\cos^3(\theta)\sin^3(\theta) and so the ... 1 The answer should be the largest of the a_i. To get to it intuitively from what you have in your last equation, note that if a>b, e^{\lambda a} grows faster than e^{\lambda b}, since e^{\lambda(a-b)} \to \infty. You can use this principle to say that the top is dominated by the term where a_i is largest, as is the bottom, so the limit tends to ... 1 Hint: Since a_1a_2=6 and a_2<a_1, either a_1=3 and a_2=2, or a_1=6 and a_2=1. 1) If a_1=3 and a_2=2, we have 27\le M<81 and 25\le M<125. 2) If a_1=6 and a_2=1, then 3^6\le M<3^7 and 5\le M<25. 1$$y-2.686=10^{1.830 \log x} \Rightarrow \log_{10}{(y-2.686)}=\log_{10}{10^{1.830 \log x}} \\ \Rightarrow \log_{10}{(y-2.686)}=1.830 \log x \Rightarrow \log x= \frac{\log_{10}{(y-2.686)}}{1.830} $$And if we suppose that the base of the logarithm is e, then it will be as follows:$$e^{\log x}=e^{\frac{\log_{10}{(y-2.686)}}{1.830}} \Rightarrow ... 1 Assumingz\geq 1$, we have: $$\int_{z-1}^{z}\log\left(\frac{1}{z-y}\right) e^{-y^3}\,dy = -\frac{d}{d\alpha}\left.\int_{z-1}^{z}(z-y)^{\alpha}e^{-y^3}\,dy\,\right|_{\alpha=0}$$ hence the original integral depends on the derivatives of incomplete gamma functions. 1 The answer is "no" as pointed out in the comments. You can however do the following:$\frac{d}{dx} \ln(f(x)) = \frac{1}{f(x)}\cdot f^{\prime}(x)$by the chain rule. So, if$\frac{d}{dx}\ln(f(x))=g(x)$then$f^{\prime}(x)=f(x)g(x)$. This is usually called "logarithmic differentiation" and tends to show up in text books near applications of derivatives or ... 1 Hint: Expand$\ln(1-u)$into its Mercator series, and then reverse the order of summation and integration. 1 Write$x^2=x^2-1+1$so that $$\frac{x^2}{x+1}=\frac{x^2-1}{x+1}+\frac{1}{x+1}=(x-1)+\frac{1}{x+1}$$ 1 We need to eliminate$yx+z-2y=x+z-4zx/(z+x)=\dfrac{(z-x)^2}{z+x}\displaystyle\log(z+x)+\log(x+z-2y)=\log(z+x)(x+z-2y)\displaystyle=\log(z+x)\cdot\dfrac{(z-x)^2}{z+x}=\log(z-x)^2$If$x>z,\log(z-x)^2=2\log(x-z)$1 I think there is a property of logarithm that you need. $$y = \log_{1/a}(x) = -\log_a (x)$$ 1 Here is an easier way out. Note that if$x \neq n\pi$, we have$\vert \cos(x) \vert < 1$. This would mean$\vert\cos(x)\vert^{\sin(x)}<1$, since$\sin(x) \neq 0$as$x \neq n\pi$. Hence, we only need to consider the case$x=n\pi$. We see that$x=n\pi$gives us that$\cos(n\pi)^{\sin(n\pi)}=1$. 1 We should take care of the case$\cos(x) = 0$separately: when$\cos(x) = 0$,$\sin(x) = \pm 1$, and$0^1 = 0$while$0^{-1}$is undefined; neither are$1$, so no solutions there. You do have$\cos(x)^{\sin(x)} = (\pm 1)^0 = 1$when$x = n \pi$. Otherwise, if you're interested in real solutions, you have$0 < |\cos(x)| < 1$and either$\sin(x) > 0\$ ...

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