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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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