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3

Hint. An elementary approach. One may consider $$ I_0=1-\frac1{e}, \quad I_n=\int_{\large\frac1{e}}^1\left(-\ln x \right)^n\:dx, \quad n\ge1, $$ then integrating by parts, $$ \begin{align} I_n&=\left[ x \frac{}{}\left(-\ln x \right)^n\right]_{\large\frac1{e}}^1 +n\int_{\large\frac1{e}}^1\left(-\ln x \right)^{n-1}\:dx \\&=-\frac1{e}+nI_{n-1} \end{...


3

$$\log _{ 2 }{ (4^{ x }-2(2^{ x })+17)>5 } \\ { 4 }^{ x }-{ 2 }^{ x+1 }+17>32\\ { 2 }^{ 2x }-2{ \cdot 2 }^{ x }-15>0\\ \left( { 2 }^{ x }+3 \right) \left( { 2 }^{ x }-5 \right) >0\\ { 2 }^{ x }\in \left( -\infty ;-3 \right) \cup \left( 5;+\infty \right) \Rightarrow x\in \left( \log _{ 2 }{ 5 } ;+\infty \right) $$ Can you take from here?


2

$1)$ Here's a solution to the first question $$\int_1^x t^t dt$$ This is an anti-derivative of the function $f(x)=x^x$. $2)$ $\ln(-1)$ is already extended by anjoining $i$, or at least in a sense. Let $z=re^{i\theta}$ for $0 \leq \theta <2\pi$. Then we can define a complex logarithm as $\text{Log}(z)=\ln(r) + i\theta$. Then $\text{Log}(-1)=\text{...


2

You have $$x^2+(4+b)x+16=0\tag1$$ This is correct. However, note that when we solve $$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ we have to have $$bx+28\gt 0\quad\text{and}\quad 12-4x-x^2\gt 0,$$ i.e. $$bx\gt -28\quad\text{and}\quad -6\lt x\lt 2\tag2$$ Now, from $(1)$, we have to have $(4+b)^2-4\cdot 16\ge 0\iff b\le -12\quad\text{or}\quad b\ge 4$. ...


2

The exact expectation for any $x$ is $$E_x = x H_x = x \sum_{n=1}^x \frac1n$$ $H_x$ is called the $x$-th harmonic number. Your expressions are both wrong, but really close; the reality lies between them: For large $x$, $$E_x = x ((\ln x )+ \gamma) + \frac12 - \frac1{12x} + \frac1{120x^3}+ O\left( \frac1{x^5}\right)$$ where $\gamma$ is Euler's constant, ...


2

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities $$\frac{x-1}{x}\le \log(x)\le x-1$$ for $x>0$. Then, letting $x\to x+1$ we arrive at the coveted inequalities $$\frac{x}{x+1}\le \log(1+x)\le x$$ And we are done! Note that we can also ...


2

We have $$a^2+b^2=c^2$$ so that $$(c+b)(c-b)=c^2-b^2=a^2.$$ Take the logarithm of both sides $$\log(c+b)+\log(c-b)=2\log(a).$$ Divide by $\log(c+b)$: $$\frac{\log(c+b)}{\log(c+b)}+\frac{\log(c-b)}{\log(c+b)}=2\frac{\log(a)}{\log(c+b)}$$ Simplify, using the fact that $\frac{\log(x)}{\log(y)}=\log_y(x)$: $$1+\log_{c+b}(c-b)=2\log_{c+b}(a).$$ Multiply by $\...


2

If $x$ and $y$ aren’t too awfully large, you can simply calculate $\log_{10}x^y=y\log_{10}x$. For example, $$\log_{10}12^{400}=400\log_{10}12\approx431.6725\;,$$ which tells you that the number has $431+1=432$ digits in base ten.


2

$log(a)+log(b)=log(a\cdot b)$ Solution: $log_2(x+2)(x-5)=3$ $\longrightarrow$ $(x+2)(x-5)=2^3$ $\longrightarrow$ $x^2-3x-10=8$ $\longrightarrow$ so the answer is x=6 and x=-3. I hope this helped.


1

You just made a mistake that $$\color{red}{\log_d a + \log_d b =c}$$ $$\color{red}{a+b=d^c}$$ $$\color{green}{\log_d a + \log_d b=\log_d a b}$$ $$\log_2(x+2)+\frac{1}{2}\log_2(x-5)^2=3$$ $$\log_2(x+2)+\frac{2}{2}\log_2(x-5)=3$$ $$\log_2(x+2)+\log_2(x-5)=3$$ $$\log_2\ (x+2)(x-5)=3$$ $$x=6$$ $\color{red}{x\ne -3}$ because it is not in domain .


1

You should have $(x+2)(x-5)=2^3$. The exponent of a sum is the product of the exponents.


1

In general, you can calculate the number of digits in an arbitrary base $n$ of an expression $a^b$ by the formula $$D = \lfloor 1 + \log_{n}(a^b)\rfloor = \lfloor 1 + (b)\log_{n}(a)\rfloor$$ where $D$ represents the number of digits in your result.


1

Let $f(x)=\ln(1+ax)$ and $x_0=0$ in the difference quotient. Then $$\frac{f(x)-f(0)}{x-0}=\frac{\ln(1+ax)-\ln(1+a(0))}{x-0}=\frac{1}{x}\ln(1+ax).$$ This tells you that the limit you want is just $f'(0)$. Proving the limit in this way (or using L'Hospital's rule as in the other answer) is dishonest/circular because you need to know the value of the limit ...


1

L'Hôpital actually works, as $\lim \log(1+ax) = \log(1+0) = 0$. Therefore $$\lim \frac{\log(1+ax)}{x} = \lim \frac{a}{1+ax} = a.$$ "$a$ is fixed" means that $a$ is a predetermined number, i.e., a constant.


1

your equation system can be written as $$\frac{28.8-24.5}{C}=e^{-kt}$$ $$\frac{28.0-24.5}{C}=e^{-kt}\cdot e^{-\frac{29}{60}k}$$ plugging (1) in (2) we obtain $$\frac{28.0-24.5}{C}=\frac{28.8-24.5}{C}\cdot e^{-\frac{29}{60}k}$$ Can you proceed?


1

You want to eliminate one of $k, t$ so from the first equation, we get $Ce^{-kt} = 4.3$, plugging this into the second equation, we get $$3.5 = Ce^{-kt}\cdot \exp\left({-\frac{29k}{60}}\right) = 4.3 \exp \left(-\frac{29k}{60}\right)$$ which you can then use to solve for $k$ easily. Note, we use the property that $e^{a+b} = e^a \cdot e^b$.


1

$ 12 ^ p = 18$ Equation 1 $ 24 ^ q = 54 $ Equation 2 (1) *3 = (2) ,You can write 12,24 and 18 as product of powers of 2 and 3 and then equate exponents on both side


1

HINT: The idea is to eliminate $\log2, \log3$ $$(2p-1)\log2=(2-p)\log3$$ $$(3q-1)\log2=(3-q)\log3$$ Divide the the first relation by the second and rearrange.


1

The change of base formula for logarithms says that $$ \log_xy=\frac{1}{\log_yx} $$ assuming $x$ and $y$ positive and different from $1$. Thus, assuming $a\ne1$, $c-b\ne1$ and $c+b\ne1$, we have \begin{align} \log_{c+b}a+\log_{c-b}a &= \frac{1}{\log_a(c+b)}+\frac{1}{\log_a(c-b)}\\[6px] &= \frac{\log_a(c-b)+\log_a(c+b)}{\log_a(c+b)\cdot\log_a(c-b)}\\[...


1

$1)$ WolframAlpha did not give me the full solution, but you can get it to give you the first 24 terms by pressing 'more digits'. $$\int x^xdx=x+\frac{2\log(x)-1}4x^2+\frac{9\log^2(x)-6\log(x)+2}{54}x^3+O(x^4)$$ $2)$ Start with $(-1)^{-1}=(-1)^1=-1$. Thus, it is sufficient enough to show $\ln(-1^1)=\ln(-1^{-1})$. In other words, $\ln(-1)=-\ln(-1)$. ...


1

This one is similar to this answer of your previous question and you should have guessed it. Based on the definition of Ramanujan's class invariants \begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\tag{1}\\ g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\...



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