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16

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

$$x = \log_2 y \iff y = 2^x$$ The 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

Define $$e=\lim_{h\to 0} \left(1+h\right)^{1/h}.$$ 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 setting $x=u^2$ 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, setting $t=\frac{\log ( x)}{\log (2)}=\log_2(x)$, ...


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 by $x+5$ since the question has the implicit constraint that $x>0 \implies x+5 > 0$. If you want to work with complex-valued ...



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