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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)=3x=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, $\ln(-1)=-\ln(-1)$. ...