107

$\DeclareMathOperator\antilog{antilog}$ First, when Napier invented logarithms, his application was not the inverse of the exponential function. He was originally multiplying quantities called "sines", which are not what you're thinking when you see that word. These "sines" are vaguely similar to the positions of the arrow in Zeno's ...


19

Context is important here. In the context of real numbers, negative numbers have no logarithms (and neither does $0$) because $\log(x)$ is a number $y$ such that $e^y=x$ and $e^y$ is always greater than $0$. On the other hand, in the context of complex numbers, every complex number other than $0$ has logarithms. In fact, any such complex number has ...


14

Consider the function $$f_p(x)=\frac{x^p-1}p.$$ It has the derivative $$f'_p(x)=x^{p-1}$$ and is such that $f_p(1)=0$ and $f_p(0)=-\dfrac1p.$ Now if you let $p$ tend to $0$, you have that $$\lim_{p\to0}f_p(x)=\ln(x)$$ and $$\lim_{p\to0}f'_p(x)=\frac1x.$$ Below, a pencil of curves for various positive and negative $p$. Also consider the inverse of this ...


13

The Euler product factorization of the Riemann zeta function is: $$ \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_p \left(1-\frac{1}{p^{s}}\right)^{-1} $$ Apply natural logarithm with the Newton-Mercator expansion $$ \ln\zeta(s)=\sum_p -\ln\left(1-\frac{1}{p^s}\right)=\sum_p \sum_{k=1}^{\infty}\frac{1}{kp^{ks}}. $$ This is the special case $s=2$.


12

Opening: To begin, whenever one finds an equation of this form to solve, they almost always have to resort to use of the Lambert W function - though a deceptively simple equation, it's actually very difficult to solve for $x$ when it is in these two different positions (as a linear term and in an exponent). Some simpler cases (such as $2^x = x+2$) can often ...


10

Hint. Let $z=e^{-x/4}$ and solve the quadratic equation with respect to $z$. $$y = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}={ 1 \over 3} z^2 + { 2 \over 3} z\Leftrightarrow (z+1)^2=1+3y.$$ P.S. As pointed out by Yves Daoust, starting from the hint we easily find $$x=f^{-1}(y)=-4\log\left(\sqrt{3y+1}-1\right)$$ which can be written also in the following ...


8

As noted, the sequence goes like $$ 0,\underset{2}{\underbrace{1,1}},\underset{4}{\underbrace{2,2,2,2}},\underset{8}{\underbrace{3,3,3,3,3,3,3,3}},4,4,\ldots$$ i.e, every natural number $k$ occurs $2^k$ times. So desired is $$ \sum k\cdot 2^k =1994$$ It's quick enough to attack directly: $$ 1\cdot2 + 2\cdot4 + 3\cdot8 + 4\cdot16 + 5\cdot 32 + 6\cdot 64 + 7\...


8

Using Sterling’s approximation for the Gamma function, $$\Gamma(z)\sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z$$ In this case \begin{align*} \Gamma\left(x+\frac{1}{2}\right)&\sim \sqrt{\frac{2\pi}{x+\frac{1}{2}}}\left(\frac{x+\frac{1}{2}}{e}\right)^{x+\frac{1}{2}}\\ \ln\Gamma\left(x+\frac{1}{2}\right)&\sim \frac{1}{2}\ln 2+\ln \sqrt \pi -\...


7

Hint $$\int_1^z \frac{1}{u \cdot \sqrt{u^2 + 1}} \,du=\int_1^z \frac{u}{u^2 \cdot \sqrt{u^2 + 1}} \,du$$ Set $v= \sqrt{u^2+1} , dv =\frac{u}{\sqrt{u^2+1}}du$. $$\int_1^z \frac{u}{u^2 \cdot \sqrt{u^2 + 1}} =\int_{\sqrt{2}}^{\sqrt{z^2+1}} \frac{dv}{v^2-1} \,du$$


7

Let $a=\ln(t)$, $b=\ln(t+1)$ and $x=1/\ln(n)$. Then $x\to 0$ as $n\to +\infty$ and $$\ln^2(n)\ln \left (\frac{\ln(\frac{n}{t})}{\ln(\frac{n}{t+1})}\right )-\ln(n)\ln \left (1+\frac{1}{t} \right)=\frac{\ln(1-ax)-\ln(1-bx)-(b-a)x}{x^2}.$$ Now by using the Taylor series $$\ln(1+y)=y-\frac{y^2}{2}+O(y^3),$$ we find that the above LHS is $$\frac{b^2-a^2}{2}+O(x)....


7

Yes, you are right. Simplify as follows $$y=\log_{\sqrt[3]{x}}(7)=\frac{\ln 7}{\ln (\sqrt[3]{x})}=\frac{\ln (7)}{\frac13\ln x}=\frac{3\ln (7)}{\ln x}$$ $$\therefore \frac{dy}{dx}=3\ln (7)\left(\frac{-1}{(\ln x)^2}\frac1x\right)=-\frac{3\ln (7)}{x(\ln x)^2}$$


7

They are correct. Anything that is $O(n^3)$ is also $O(n^4)$.


7

Hint: After cheating with WA, here is the solution: notice that "by magic", $$\int\frac{x^2(1-\log x)}{(\log x)^4-x^4}dx =\int\frac{\dfrac{1-\log x}{x^2}}{\dfrac{(\log x)^4}{x^4}-1}dx =\int\frac1{\dfrac{(\log x)^4}{x^4}-1}d\frac{\log x}x.$$ Now $$\frac1{u^4-1}=\frac1{4 (u-1)} -\frac 1{4 (1 + u)} - \frac1{2 (1 + u^2)}$$ which is easy to integrate.


7

$\displaystyle 9^{\log_3 n} = 3^{2\log_3 n} = (3^{\log_3 n})^2 = n^2$


7

No. For example, $\log 2+\log9=\log 18<\log 25=\log5+\log5$, though $2+9=11>10=5+5$.


7

The map $$f_x(y) = \frac{y-x}{\ln(y)-\ln(x)}$$ defined on $(x, \infty)$ is continuous and positive. As it is not constant, it takes rational values according to the intermediate value theorem as $\mathbb Q$ is dense in $\mathbb R$.


7

In general it is not possible to take a number $x$ and a negative number $n$ and find an exponent $y$ such that $n^y=x.$ For example, it won't work if $x=7$ and $n=-7.$ It works out in the specific case of $x=49$ and $n=-7$, but it's just not useful enough to define the logarithm function for a few special values where logarithm by a negative number actually ...


7

This is because the logarithm function with a specific base is injective. For an injective function $f$, it holds that if $f(x) = f(y)$, then $x=y$. You can prove this in many ways, for example that it is the inverse function of the exponential function.


6

Since COVID-19 is a viral disease, the number of people who get infected grows exponentially (until the curve hits the inflection point). Since exponential graphs grow really fast, its hard to study data from the graph normally. When you graph them on a logarithmic scale, $\log a^x$ grows linearly since: $$\log a^x=cx$$ for $c=\log a$. This allows us to ...


6

The problem, as you have written it, has no solution. Simplifying the LHS, we get $$(a^2+1)(b^2+1)(c^2+1) = 4abc$$ But, by the AM-GM inequality, we get $x^2+1\ge2x$, which gives $$(a^2+1)(b^2+1)(c^2+1) \ge 8abc$$


6

No you can't find a bijection on $\mathbb{R}$ with this property, the only function is $f\equiv 0$. Indeed you have $f(0) = f(0\times x) = f(0) + f(x)$ so that $f(x) = 0$. This is true for all $x\in\mathbb{R}$. If you exclude the $\{0\}$ again you cannot have a bijection since you have that $f$ must be a even function. Indeed $f(x)=\frac{1}{2}f(x^2)=f(-x)$. ...


6

Because $2x-3y=1$, we know that $x = \frac{1+3y}{2}$. Then $2^x+3^y = 7$ becomes $$2^{(1+3y)/2}+3^y=7$$ so it suffices to solve the above equation. Observe that the function $$f(y) = 2^{(1+3y)/2}+3^y$$ is increasing (exponential functions with base $>1$ are increasing, and the sum of two increasing functions is increasing). In particular, it is a one-to-...


6

The indefinite integral actually equals $$\ln \color{red} |e^{2x}-1\color{red} |-x+C=\begin{cases} \ln(e^{2x}-1) -x+C_1,& x\ge 0 \\ \ln(1-e^{2x})-x+C_2, & x\lt 0 \end{cases}$$


6

Here is a partial proof. I have some ideas, but it will be a while before I can return to it. I'll show that the conjecture is true for $m<\sqrt{\log{(\pi/2)}} \ n \sim .672 \ n.$ Maybe someone else can use these ideas for a full proof. Use the fact that the central binomial $\binom{2n}{n+m}$ has its max at $m=0.$ Do an asymptotic expansion $$ \binom{...


6

With @skbmoore's work, we know that this is true for $m<\sqrt{\log(\pi/2)}n$. I'll now show that it's also true for $m>\frac12n$, which will obviously prove the result. Rearrange the desired inequality as thus: $$\log\left(\frac{4^n}{\sqrt{2n+1}}\right)\ge\frac{m^2}n+\log\binom{2n}{n+m}=f_n(m).$$ We're basically going to try to show that $f_n(m)$ is ...


6

Remember $(a^b)^c=a^{bc}=(a^c)^b$, so $$ (2^3)^{\log_2 n}=(2^{\log_2 n})^3=n^3. $$


6

The solution is $$ x = \sqrt{\frac{W(1572864 \ln(2))}{6 \ln(2)}} $$ where $W$ is the Lambert W function. EDIT: To see this, let $y = 6 x^2 \ln(2)$, and rewrite the equation as $$ \frac{e^{y/6}}{8} = \frac{2^{y/(6 \ln 2)}}{8} = y^{-1/6} (6 \ln(2))^{1/6}$$ Taking the $6$'th power of each side and multiplying by $8^6 y$, it becomes $$y e^y = 1572864 \ln(2)$$ so ...


6

The expression $\ln(x/y)$ only makes sense when $x/y>0$. In this specific case we have, $x>0$ and $y>0$, or, $x<0$ and $y<0$. When ($x>0$ and $y>0$) or ($x<0$ and $y<0$), we can write, $$\frac xy=\frac{|x|}{|y|}$$ and then, $$\ln (x/y)=\ln |x|-\ln |y|.$$


6

Here is an intuitive reason for why this is true. Note that $\ln|x|, x < 0$ is a reflection of $\ln|x|, x > 0$ through the $y$-axis. Now recall that the instantaneous rate of change $\frac{dy}{dx}$ is loosely speaking, $\frac{\text{change in }y}{\text{change in }x}$ for a very small interval. The change in $y$ is unaffected, but when $x < 0$, the ...


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