5

$$\log_{17}291 > \log_{17}289 = 2$$ $$\log_{17}291 > 2$$ $$\log_{13}160 < \log_{13}169 = 2$$ $$\log_{13}160 < 2$$ Thus: $$\log_{13}160 < 2 < \log_{17}291$$ $$\log_{13}160 < \log_{17}291$$


4

For $n <t<n+1$ we have $\frac 1 {n+1}<\frac 1t <\frac 1 n$. Integrate from $n$ to $n+1$ and see what you get.


3

For any real $w$, by the Taylor formula with Lagrange remainder, we have $$ e^{w - 1} = 1 + (w - 1) + \frac{{(w - 1)^2 }}{2}e^\xi \ge 1 + (w - 1) = w $$ with some $\xi$ between $0$ and $w-1$. Thus, for any $w>0$, $w-1 \ge \log w$. With $w=x/x_0$, where $x$, $x_0>0$, we get $$ \log \left( {\frac{x}{{x_0 }}} \right) \le \frac{x}{{x_0 }} - 1 \...


2

With $y=e^x$ render $y^2+y^{-3}+y^4-y^{-5}-1=0$ $y^7+y^2+y^9-1-y^5=0$ $\color{blue}{y^9}+y^7-y^5+y^2\color{blue}{-1}=0$ The leading term and the constant have opposite signs for positive $y$, so there must be a positive root for $y=e^x$, forcing a real root for $x$.


2

Define $h(x):=x-1-\frac{\ln x}{\ln 10}.$ Note $h'(x)=1-(x\ln 10)^{-1}$ and $$h'(x)\lesseqqgtr 0\iff x \lesseqqgtr (\ln 10)^{-1}.$$ Note $$\lim_{x\rightarrow 0^+}h(x)= +\infty, \\ h((\sqrt 10)^{-1})=(\sqrt 10)^{-1}-2^{-1}<0,$$ so IVT implies $\exists c\in (0,(\sqrt 10)^{-1})$ s.t. $h(c)=0.$ From the above you should be able to convince yourself that $$x-1&...


2

Hint: $\log x$ is concave, so it remains below its tangent at $x_0$.


2

Let $a:=x/x_0 >0$, where $x, x_0 >0$. Want to show $\log a \le a-1.$ Recall $\displaystyle{\int_{1}^{a}}(1/t)dt =\log a;$ 1)$a\ge 1:$ $\log a = \displaystyle{\int_{1}^{a}}(1/t)dt \le \displaystyle{\int_{1}^{a}}(1)dt,$ $\log a \le a-1.$ 2)$1>a>0:$ Then $\displaystyle{\int_{a}^{1}}(1/t)dt >\displaystyle{\int_{a}^{1}}(1)dt=1-a;$ $(-1)\...


2

A solution idea of calculating the integral by Cornel Ioan Valean The post is extremely short since I have no time, but once you know what to do, all is trivial. So, what do to? Observe that any $J(2n,k)$ is half the real part of the integral over the positive real line. $$\int_{0}^{1} \frac{\log(x)^{2n}\log\left(\frac{1-x}{1+x}\right)}{(x-1)^2-k^2(x+1)^2}\...


1

If $|x+1|\ge 0$ the equation becomes $$ 2\cdot 2^x-|2^x-1|=2^x+1 $$ that is equivalent to $$ \begin{cases} 2^x-1\ge0\\ 2\cdot 2^x-2^x+1=2^x+1 \end{cases} \quad \lor\quad \begin{cases} 2^x-1<0\\ 2\cdot 2^x+2^x-1=2^x+1 \end{cases} $$ can you solve these two systems? and solve also the other case: for $|x+1|< 0$ ?


1

For $x\geq0$ the equation becomes $$2^{x+1}-(2^x-1)=2^x+1$$ and the left hand side simplifies to $2^x+1$ so that every $x\geq 0$ is a solution. For $x<0$ the equation becomes $$2^{|x+1|}+(2^x-1)=2^x+1$$ so we get $2^{|x+1|}=2$ so that $|x+1|=1$ so that $x=-2$ is another solution.


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