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The simpler way is to use the inverse function theorem for derivatives: If $f$ is a bijection from an interval $I$ onto an interval $J=f(I)$, which has a derivative at $x\in I$,and if $f'(x)\neq 0$, then $f^{-1}\colon J\to I$ has a derivative at $y=f(x)$, and $$\bigl(f^{-1}\bigr)'(y)=\frac1{f'(x)}=\frac1{f'\bigl(f^{-1}(y)\bigr)}.$$ As $(\mathrm ... 11 If you can use the chain rule and the fact that the derivative of$e^x$is$e^x$and the fact that$\ln(x)is differentiable, then we have: $$\frac{\mathrm{d} }{\mathrm{d} x} x = 1$$ $$\frac{\mathrm{d} }{\mathrm{d} x} e^{\ln(x)} = e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$ $$e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$ x ... 8 In the First : \sinh x = \frac{e^x - e^{-x} }{2} \Longrightarrow \sinh^{-1}x = \ln(x+\sqrt{x^2+1}). to get this just solve the equation y=\sinh x to get the above inverse function (notice that e^x>0). The integral becomes : \begin{align*} \int_0^{2\pi} \sinh^{-1} \sin x\ \mathrm{d}x &= \int_0^{\pi} \sinh ^{-1} \sin x \ \mathrm{d}x + ... 7 Here is a 'real-analysis route'. Step 1. We have \int_0^{+\infty}\frac{\ln x}{x^2+1} dx=0 \tag1$$as may be seen by writing$$ \begin{align} \int_0^{+\infty}\frac{\ln x}{x^2+1} dx&=\int_0^1\frac{\ln x}{x^2+1} dx+\int_1^{+\infty}\frac{\ln x}{x^2+1} dx\\\\ &=\int_0^1\frac{\ln x}{x^2+1} dx-\int_0^1\frac{\ln x}{\frac{1}{x^2}+1} \frac{dx}{x^2}\\\\ ... 6 Let\log_2 x = t$. Then $$\log_{1/2} (4x) = \frac{\log_2(4x)}{\log_2(1/2)} = -(2+t)$$ $$\log_2\frac{x^2}{8} = 2t-3$$ So we solve$(2+t)^2+2t-3=8 \iff (t-1)(t+7)=0$, or$x=2, \dfrac1{2^7}$for a product of$\frac1{64}$. 5 You can't exponentiate both sides just yet (well, you can, but I'd rather not), let's see what you can do instead using$2 \ln x = \ln x^2$giving us $$\ln x^2 = \ln (x^2 + x -3).$$ Now you can raise$e$to the power of each side (exponentiate each side) and get$x^2 = x^2 + x - 3$which is solvable and gives$x = 3. Let's see if this works: $$e^{2 \ln 3 ... 4$$x = \log_2 y \iff y = 2^xThe inverse is called the base two logarithm. In your case 2^x = 8 \iff x = \log_2 8 = \log_2 2^3 = 3. In general the inverse for a^x, where a> 1 is the base a logarithm. So y = a^x \iff x = \log_a y. 4 Here is how to do it using complex analysis. First of all in this case you can't compute \frac{1}{2}\int_{-\infty}^\infty \frac{\ln x}{(x^2+1)^2} since it does not equal your integral (why?). Now take R>1, r<1 and \gamma a "keyhole" contour as shown below Lets take the branch cut of the logarithm with domain \mathbb{C} \setminus \{x+iy: y ... 4 As far as I know, this doesn't have an elementary solution. The best I can get using numerical methods is x \approx 5.94051. Was this an exercise in numerical methods? If you meant \log_2 x - 3, then it can be solved the following way: \begin{align} \log_2 x \cdot (\log_2 x - 3) &= 4 \\ (\log_2 x)^2 - 3 \log_2 x - 4 &= 0 \end{align} ... 4 Definee=\lim_{h\to 0} \left(1+h\right)^{1/h}.$$Then change variables h\mapsto h/x giving$$e=\lim_{h/x\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}}=\lim_{h\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}},$$where the limit in the second equality follows since h approaches 0 as h/x does. Since x is constant w.r.t. h, we can simplify by ... 4 No, you should prove that the limit exists and finite in the first place. This is a common error that students make. How to prove that the limit exist and finite? 1) We prove by induction that a_n\in (0,2) for any n\in \mathbb{N} : if 2>a_n>0 then$$2=\sqrt{2+2}>\sqrt{2+a_n}=a_{n+1}=\sqrt{2+a_n}>0.$$and clearly a_n\in (0,2). 2) Now, ... 4 You don't have to pass through the exponential.$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \sqrt{2 + a_{n-1}} = \sqrt{2 + \lim_{n\to\infty} a_{n-1}} = \sqrt{2 + \lim_{n\to\infty} a_{n}} .$$Let A = \lim_{n\to\infty} a_n, then$$ A = \sqrt{2 + A}$$So A = 2. P.S. @aziiri is right. To complete the proof you have to prove that the limit exists and ... 3 I really don't know how to start solving this problem This is hardly surprising, since the anti-derivative cannot be expressed in terms of elementary functions. any tips or solutions will be greatly appreciated. Add some integration limits:$$\int_0^{\big(\tfrac\pi2\big)^2}\ln\Big(\sin\big(\sqrt{x}\big)+\cos\big(\sqrt{x}\big)\Big)~dx ~=~ ... 3 By settingx=u^2$we have: $$I = \int 2u\log(\sin u+\cos u)\,du = u^2\log(\sin u+\cos u)-\int u^2\frac{1-\tan u}{1+\tan u}\,du$$ and, by putting$v=\frac{\pi}{4}-v$, $$-\int u^2\frac{1-\tan u}{1+\tan u}\,du = \int \left(\frac{\pi}{4}-v\right)^2 \tan v\,dv$$ Now we may exploit$\int\tan v\,dv=-\log\cos v , so the last integral just depends on: $$\int v ... 3 Go ahead and take the \log_{10} on both sides:$$3\log_{10}(x)-\log_{10}(x)^2+\log_{10}(x)\log_{10}(3)=\log_{10}(900).$$Now solve the quadratic. Let y=\log_{10}(x). Then this quadratic is$$y^2-(\log_{10}(3)+3)y+\log_{10}(900)=0.$$Applying the quadratic formula, we get ... 3 e^{x/e} is convex and its tangent at x=e is y = x, hence e^{x/e} \ge x 3 Not necessarily. Consider Y such that Pr[Y=0]=Pr[Y=1]=1/2. Define g(x,Y)=e^{Yx}. Then g(x,Y) is log concave in x because \log g(x,Y) = Yx is linear. But:$$ E[g(x,Y)] = \frac{1 + e^x}{2} $$and \log E[g(x,Y)] = \log(1/2) + \log(1 + e^x), which is no longer concave. 2 You are alsmot there. Study the function f(x)=\dfrac{\ln x}{x} Then f'(x)=\dfrac{1}{x^2}(1-\ln x) f'(x)>0 for x<e and f'(x)<0 for x>e But f(e)=\dfrac 1e... 2 Let the three expressions be equal to a. log_{15}\frac{2}{9} = log_{3}x = log_{5}(1-x) = a It follows that \frac{2}{9} = 15^a , x = 3^a and 1-x = 5^a \frac{2}{9} = 15^a = 3^a \times 5^a Hence, x \times (1-x) = \frac{2}{9}, Can you solve this now? x(1-x) = \frac{2}{9}, 9x^2 - 9x +2 = 0 Hence, x = \frac{1}{3} or x = \frac{2}{3} ... 2 If you can use the definition of e as:$$e:=\lim_{n\rightarrow∞}\left(1+\frac{1}{n}\right)^n$$and the slightly modified form: \displaystyle e^x=\lim_{n\rightarrow∞}\left(1+\frac{x}n\right)^n then, by setting h=\frac1{x} you can calculate the desired limit. 2 As the exponential function is strictly growing, the equation 2^x=y has a single real solution in x (for y>0), which is called the logarithm (in base 2), denoted as \log_2(y).$$2^{\log_2(y)}=y.$$The values of 2^x for increasing integers x are 1,2,4,8,16,32\cdots, and for negative integers, \dfrac12,\dfrac14,\dfrac18\dfrac1{16}\cdots, ... 2 2 \ln(x)= \ln(x^{2}+x-3) so \ln(x^{2})=\ln(x^{2}+x-3) so x^{2}=x^{2}+x-3 so x=3. Added later : Notice that we must have x>0 and x^{2}+x-3>0, so 3 is an acceptable value!. 2 Here is another way of evaluating the integral. Let f(x) be an odd function with period 2\pi. Then, we will show that$$\int_0^{2\pi}\log \left(f(x)+\sqrt{f^2(x)+1}\right)dx=0\begin{align} \int_0^{2\pi}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx&=\int_{-\pi}^{\pi}\log\left(f(x) +\sqrt{1+f^2(x)}\right)dx\\\\ &=\int_{-\pi}^{0}\log\left(f(x) ... 2 Going to natural logarithms (the only I know, if I may confess), you have $$\log_{1/2}(4x)=-\frac{\log (4 x)}{\log (2)}=-2-\frac{\log ( x)}{\log (2)}$$ $$\log_2(\frac{x^2}{8})=\frac{\log \left(\frac{x^2}{8}\right)}{\log (2)}=\frac{2\log ( x)}{\log (2)}-3$$ So, settingt=\frac{\log ( x)}{\log (2)}=\log_2(x)$, ... 2 Usually,$\log$means$\log_e = \ln$or$\log_{10}$. Either way, there isn't a neat answer to the question. The most you can do is write$\ln 64 = 6\ln 2$and$\log_{10}64 = 6 \log_{10}2$, using that$2^6=64$. Now the problem boils down to knowing the values of$\ln 2$and$\log_{10}2$. These values can be approximated numerically using calculus, for ... 2 Because the derivative of$\ln(f(x))$is not$\frac{1}{f(x)}$for all differentiable function$f$, even if it is true for$f(x)=x-a$where$a$is a constant. The derivative of$\ln(f(x))$is$\frac{f^\prime(x)}{f(x)}$applying the chain rule. 1 The chain rule is the difference. Note that$\int\frac{du}{u}=\ln|u|$. So, you must have a fraction of the form$u$on the bottom and the derivative of$u$on the top. For your second example,$u=1-x^2$, but$du=-2xdx$is not the numerator. 1 As already said in comments and answers, this kind of equation cannot be solved in terms of elementary functions. As imulsion showed, there is a analytical solution in terms of Lambert function and for$ax+b^x=c$, the solution will be $$x=\frac{c}{a}-\frac{W\left(\frac{\log (b) b^{\frac{c}{a}}}{a}\right)}{\log (b)}$$ Otherwise, only numerical methods will ... 1 Start by following your nose: $$\log_2(\log_2 x) = \log_2\left( \frac{\ln x}{\ln 2}\right) = \log_2(\ln x) - \log_2(\ln 2) = \frac{\ln(\ln x)}{\ln 2} - \log_2(\ln 2) \geq \lfloor \ln(\ln x) \rfloor$$ iff $${1 \over \ln 2} \left(\ln(\ln x) - \ln 2 \cdot \lfloor \ln(\ln x) \rfloor\right) \geq \log_2(\ln 2) = \frac{\ln(\ln 2)}{\ln 2}$$ iff $$\ln(\ln x) - ... 1$$x\ln \left(x\right)+5\ln \left(x\right)-5x-25 =0 \iff (x+5)\ln x - 5x - 25 = 0$Then \begin{align} (x+5)\ln x &= 5x + 25 \\ (x+5)\ln x &= 5(x+5) \\ \ln x & = 5 \iff x = e^5 \end{align} We can divide through byx+5$since the question has the implicit constraint that$x>0 \implies x+5 > 0\$. If you want to work with complex-valued ...