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10

Hint: The top is the derivative of the bottom. So let the bottom be $u$.


9

Hint: $$\log_{10}12^{300}=300\log_{10}12=300(\log_{10}2+\log_{10}2+\log_{10}3)$$


7

The function $f(x) = x - \log x$ has derivative $f'(x) = 1 - \frac1x$ which is negative for $x<1$ and positive for $x>1$. Therefore, $f$ is strictly decreasing for $x<1$ and strictly increasing for $x>1$. Since $f(1)=1$ we have $f(x)>1$ for $x<1$ and $x>1$. Therefore, $x=1$ is the only solution to $f(x)=1$.


3

By setting $\frac{1}{1+x}=z$, you are simply looking for a root of $z^a+z^b+z^c=2$.


3

$I(5)$ doesn't seem to have a closed form that I could find. On the other hand, I found numerically that $$ \begin{eqnarray}I(4) &=& -\tfrac{32}{3} \text{Li}_4(\tfrac{1}{2})+\tfrac{52}{3} \zeta(4)+\tfrac{8}{3} \zeta (2) \log^22-\tfrac{4}{9}\log^4 2. \end{eqnarray}$$ Case 3. The way to do the integral $I(3)$ is to write $$ \mathrm{li}(x) = ...


2

If you look for a continuous function $$f(x)=\log(x!)-x$$ and search for the $x$ such that $f(x)=0$, I do not think that there is any analytical solution (beside the trivial solution $x=0$). If you plot the function, you should notice that there is a root close to $x=5$ and numerical method should be used, such as Newton. Starting from a "reasonable" guess ...


2

There is also an infinite set of complex solutions (if you already had complex numbers) if you consider different branches of the complex logarithm (Where $k$ denotes the branch): $$\log(k, x) = x-1$$ $$x=e^{x-1}$$ $$xe^{1-x}=1$$ $$-xe^{-x}=-\frac{1}{e}$$ $$x = -W_k(-\frac{1}{e}) \land k \in \mathbb{Z}$$ where $W_k$ is the $k$-th branch of the $W$-Function. ...


2

Hint: try with branch of logarithmic function.


2

Let $a\in G$. We need to show that $h$ is differentiable at $a$. As $G$ is open there exists a disc $D(a,r)\subset G$. Now as $f$ does not vanish in $G$, and hence in $D(a,r)$, we define the function $$ F(z)=\int_{[a,z]}\frac{f'}{f}. $$ Then $F$ is holomorphic in $D(a,r)$ and the product $\exp(-F(z))f(z)$ is constant in $D(a,r)$, as $$ ...


1

Hint: Use the definition $$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h} \\=\lim_{h\to 0}\frac{a^h-1}{h}\\=\lim_{h\to 0}\frac{e^{\ln a^h}-1}{h}\\=\lim_{h\to 0}\frac{e^{h\ln a}-1}{h}\\=\lim_{h\to 0}\frac{1+h\ln a+\cdots-1}{h}=\ln a$$


1

First use the commonly known limit "$\lim \frac{e^h-1}{h}$", that is: $$1 = \lim_{x \rightarrow 0} \frac{e^x - 1}{x} \\ = \lim_{\log(a)x \rightarrow 0} \frac{e^{\log(a)\cdot x} - 1}{\log(a)\cdot x} \\ = \lim_{x \rightarrow 0} 1/\log(a)\cdot \frac{a^x - 1}{x}$$ and so $f'(0) = \log(a)$ by the definition of the derivative.


1

As gmath shows, you can never take the logarithm of both sides in view of proving asymptotic estimates. More explicitly, you cannot deduce (say) $f = O(g)$ from $\log f = O(\log g)$. Stated differently, if $f = O(g)$ then it doesn't follow that $e^f = O(e^g)$. Even more simply, it is not the case that $e^{O(f)} = O(e^f)$. The root cause of this is the ...


1

To expand on @Did's answer: the Lambert $W$-function is the inverse of $y = xe^x$, $x \ge -1$. Thus $x = W(y)$ if and only if $y \ge -\frac 1e$ and $y = xe^x$. Let $2 \ln x = \sqrt x$ and let $t = \ln x$. Then $2t = \sqrt{e^t} = e^{t/2}$, so $-\frac 12 te^{-t/2} = -\frac 14$. There are two solutions to this equation, but the solution $t$ with $-\frac t2 ...


1

With differentiation, you have many techniques. You are free to use whichever technique you are most comfortable with to find the derivative because all techniques give the same result. It would be acceptable to find the derivative in any of the ways you mentioned. You could even use the product rule or the limit definition if you so choose. That problem ...


1

A nice (and popular) application of a logarithmic derivative $(\log(f)'=\frac{f'}{f}$ is the formula $$ -\frac{\zeta'(s)}{\zeta(s)}=\sum_{n\ge 1}\Lambda(n)n^{-s} $$ for the Riemann Zeta function, with $\Re(s)\ge 1$, and the von Mangoldt function $\Lambda$. For a proof see here.


1

Suppose that $x>1$. Then $$\log x=\int_1^x t^{-1}dt <\int_1^x dt = x-1$$ Now suppose $0<y<1$. Then $$\log y=\int_1^y t^{-1}dt >\int_1^y t^{-2}dt=1-1/y$$ Now set $y=x^{-1}, x>1$ to get $\log x < 1-x$ on $0<x<1$. Equality is clear at $x=1$. In particular we have proven that for $x>0$ $$1-x^{-1}\leqslant \log x\leqslant x-1$$ ...


1

Let us start with the function $$f(x)=\frac{1}{(1 + x)^a} + \frac{1}{(1 + x)^b} + \frac{1}{(1 + x)^c}-2=0$$ As said before, a first change of variable $y=\frac{1}{1+x}$ allows to rewrite it as $$f(y)=y^a+y^b+y^c-2=0$$ which could already be handled easily; but we can make it nicer looking at the exponents ...


1

This is not true. If your branch is defined over all $z$, with $\Re z>0$, but IT IS NOT defined at $$ z_0=r\,\mathrm{e}^{i\vartheta}, \quad \vartheta\in (\pi/2,3\pi/2), $$ then $w=r^{1/3}\,\mathrm{e}^{i\vartheta/3}$ has positive real part, and your branch is not defined on $z_0=w\cdot w\cdot w$.


1

Since $\log_ab=\frac{\ln b}{\ln a}$, where $\ln $ is the natural logarithm. So $\log_{4n} 40\sqrt{3} = \log_{3n} 45$ can be written as $$\log_{4n} 40\sqrt{3}=\frac{\ln 40\sqrt{3}}{\ln{4n} } = \log_{3n} 45=\frac{\ln 45}{\ln{3n} }$$. Then $$\frac{\ln 40\sqrt{3}}{\ln{n}+ \ln{4}}=\frac{\ln 45}{\ln{n}+ \ln{3} }.$$ Hence we can easily get $\ln n$ and then get n. ...


1

What is in the argument of $\log_3$? It is $x^2 - 1$, right? So you need the values of $x$ such that: $$x^2 - 1 > 0$$ We have: $$x^2 - 1 > 0 \implies x^2 > 1 \implies |x| > 1$$ You got it wrong there because $\sqrt{x^2} = |x|$, so you let escape a few possibilities. Now, drawing the real line, it is easy to see that $|x| > 1$ happens if and ...


1

$x^2>1 \iff (x<-1$ or $x>1$) Alternatively, look at the graph of $x^2-1$ and see where it is above the $x-$axis To answer your other question, logarithms are exponents. $\log_3(x)$ is the exponent $y$ which makes $3^y=x$. Now if $x$ were negative, say $x=-1$, then what is the exponent $y$ such that $3^y=-1.$ Of course there is none. So $\log_3(-1)$ ...


1

If you're more careful, you can see that $d$ can be arbitrary. First of all, to solve $xy' + 2y = 0$ as you have done, you need to rewrite it as $$\frac{y'}{y} = -\frac{2}{x},$$ but that assumes $y$ is non-zero. Treating the $y = 0$ case separately, you see that $y \equiv 0$ is a solution which can be rewritten as $$y = \frac{d}{x^2}$$ where $d = 0$. As ...



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