Tag Info

Hot answers tagged

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}$$2r$sin($\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

Assuming $z\geq 1$, 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 $y$ $x+z-2y=x+z-4zx/(z+x)=\dfrac{(z-x)^2}{z+x}$ $\displaystyle\log(z+x)+\log(x+z-2y)=\log(z+x)(x+z-2y)$ $\displaystyle=\log(z+x)\cdot\dfrac{(z-x)^2}{z+x}=\log(z-x)^2$ 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 $\sin(x) > 0$ ...



Only top voted, non community-wiki answers of a minimum length are eligible