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This equation cannot be solved using “traditional” algebraic manipulations. In this case, one would use the Lambert W function: $$W(x): x = W(x)\cdot e^{W(x)}$$ or in other words, it is the solution of the equation $x = w e^w$. With this knowledge, we can try to substitute $y:=\frac{1}{x}$: $$\Rightarrow 0 = e^y-\frac{1}{y} \Rightarrow \frac{1}{y} = e^y ... 4 Prove that:$$\log_{10}999^{999}+\log_{10}2<\log_{10}1000^{999}=2997$$In other word:$$\log_{10}2<\log_{10}1000^{999}-\log_{10}999^{999}=\log_{10}\left(\frac{1000}{999}\right)^{999}$$so:$$2<\left(\frac{1000}{999}\right)^{999}=\left(1+\frac{1}{999}\right)^{999}$$It's true by Bernoulli's inequality. Next we should prove 3 \cdot ... 3 Factoring, we get:$$x^2+x+5=x^2(1+1/x+5/x^2).$$Using the above and log rules:$$\ln(x^2+x+5)=\ln(x^2)+\ln(1+1/x+5/x^2)=2 \ln(x)+\ln(1+1/x+5/x^2).$$Similarly:$$\ln(x^8-x+3)=\ln(x^8)+\ln(1/x^7+3/x^8)=8 \ln(x)+\ln(1+1/x^7+3/x^8).Can you take it from here? 3 \begin{align} e^{-4\ln\left(x\right)+8\ln\left(y\right)+2}&=e^{-4\ln x}e^{8\ln y}e^2\\\\ &=e^2\frac{\left(e^{\ln y}\right)^8}{\left(e^{\ln x}\right)^4}\\\\ &=\displaystyle\boxed{\displaystyle\frac{e^2y^8}{x^4}} \end{align} 3 We have that:\log_{16}{32} = 1.25 = \frac{5}{4}$$Notice that this can also be represented as: 16^{5/4} = 32 Why? Well, first represent 16 as 2^4 It then follows that: 16^{5/4} = (2^4)^{5/4} = 2^5 = 32 3 The two functions are inverse one of the other. When they intersect in inly one point, they do it in a point where the slope of the graph of the functions is 1. To find b solve the equations$$ b^x=\log_bx,\quad(\log b)\,b^x=\frac{1}{(\log b)\,x}=1. $$The solution is b=e^{1/e}, x=e. 3 Using any logarithm \log, we have$$\log n = \log (d^m) = m \log d,$$so$$m = \frac{\log n}{\log d} = \log_d n.$$2 \log_{16} 32=1.25 because 16^{1.25}=32 Recall that in general, \log_{a} b=c means that a^{c}=b Addendum: If you are asking how to determine what \log_{16}32 is, we first change it into exponent form as follows: \log_{16}32=x means 16^x=32. We then take the log of both sides. \log 16^x=\log 32 Then using the property of logs that lets ... 2$$\log_3x = \frac{\log_3 n}{\log_3 2} = 1.59 \log_3 n$$Because \frac{1}{\log_3 2} = 1.59 and use that c\log_a b = \log_ab^{c}. 1 Assuming that there is an integer r that works, then N is slightly smaller than 2^r, which means that \log_2 N is slightly smaller than r. \lfloor \log_2 N +1\rfloor is the next integer up. They're presumably not using \lceil \log_2 N\rceil since that gives the wrong answer in the lowest cases. 1 I think the LHS =\dfrac{\tan^{-1}\dfrac ba}\pi Let a=r\cos\theta, b=r\sin\theta where r\ge0 Square & add to get r and \dfrac ab=\cdots \implies\dfrac{a+ib}{\sqrt{a^2+b^2}}=\cos\theta+i\sin\theta=e^{i\theta} Also, e^{i\pi}=\cos\pi+i\sin\pi=-1\implies\ln(-1)=? 1 After our comment conversation, we see that the equation would be$$\text{amount}=\text{initial}(0.979)^x$$And to see how many cycles it takes to get to half the initial amount would be$$\frac{c}{2}=c(0.979)^x\\\frac{1}{2}=0.979^x$$1 Let the initial volume of the container be V_0 and the density be \rho. Let the evaporation and condensation be uniform and that 2.1% of the volume is lost everytime the purifying process is over. Thus the model is$${\rho\times(\dot V_0 - \dot V_1)} = 0.021*\rho\times\dot V_0$$Cancelling \rho, and converting the volumetric rate to volume, You ... 1 One can get a closed form for an upper (by that meaning right endpoint) Riemann sum R_n by using, instead of the partition of [1,a] into n equal parts, the partition$$x_0=a^{0/n}=1,\ x_1=a^{1/n},\ \ldots \ x_n=a^{n/n}=a.$$Then the kth subinterval [x_{k-1},x_k] has length a^{(k/n)}-a^{(k-1)/n}, and for a right endpoint sample point at x_k of ... 1 In the complex numbers, you can take the logarithm of negative numbers as you are thinking. Unfortunately, the answer is not unique because of the periodicity of \sin and \cos. From e^{i\theta}=\cos \theta + i\sin \theta you also get e^{i(\theta+2\pi)}=\cos \theta + i\sin \theta, so you can add any integral multiple of 2\pi i to the log and get ... 1 In general, \log_{a^n}(a^m) = \frac{m}{n}. One way to see this result is the general rule:$$\log_x y = \frac{\log_a y}{\log_a x}$$I your case, a=2, m=5 and n=4. This is the only way for \log_{x}y  to be rational if x,y are integers greater than 1. This is because if x=y^{p/q}, with p,q relatively prime integers, then let ... 1 Using m\log a=\log (a^m) and \log b-\log c=\log\dfrac bc where all the logarithms remain defined unlike 2\log(-1)=\log(-1)^2=\log1=0 We have \log \left(m^{.75}\right)=\log\dfrac{1050}{73.3}=\log\dfrac{1050\cdot3}{220}$$\implies m^{.75}=\dfrac{315}{22}\implies m=\left(\dfrac{315}{22}\right)^{\dfrac43}$$1 I assume, you evaluate this at values t=2\pi and t=0. Then you see immediately, that e^{2\pi i}=e^0=1 which means your expression vanishes. To evaluate the complex logarithm, the simplest way is to make the argument a complex number and then use \log z = \log |z| + i \arg z + 2ik\pi, k\in \mathrm{Z}. Here z=\frac{1-2i}{1+2i}=-\frac{3+4i}{5} \ ... 1 Let$$f(x) = \biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.$$Then$$ \begin{align*} \ln f(x) &= x\ln x[\ln \ln (x-2) - \ln \ln (x-1)] \\ &= x \ln x\left[\ln \left(1 + \frac{\ln(1 - 2/x)}{\ln x}\right) - \ln \left(1 + \frac{\ln(1 - 1/x)}{\ln x}\right) \right]\\ &= x \ln x\left[\ln \left(1 - \frac{2}{x \ln x} + o \left(\frac{1}{x\ln x} ...

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Try using logarithms. A logarithm is defined as follows, If $a^{b}=c$, then $\log_{a} c=b$ So, similarly, here we get $m=\log_{d} n$. We can further simplify it by changing the logarithms' base to $10$, $m=\frac{\log_{10} n}{\log_{10} d}$ If you want to learn more about logarithms, go here

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Hint: Unchain the various operations from each other: $$\log_4 q = -1\\ r^2 = q\\ \sin A = r$$ and finally, $A$ is an acute angle. Different direction hint: Taking the rule $\log a^b = b\log a$, we can start with $\log_4(\sin^2 A) = -1 = 2\log_4(\sin A)$.

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Hints: $$\log(a^b) = b \log(a) \\ \log(ab) = \log(a) + \log(b)$$ Example for part $c)$: $$210 = 40(1.5)^x \\ \log(210) = \log(40 (1.5)^x) \\ \log(210) = x\log(1.5) + \log(40) \\ \frac{\log(210) - \log(40)}{\log(1.5)} = x \\ x \approx 4.0897$$

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Suppose that $f(1) = 1, f(10) = 2, f(100) = 3.$ Let's suppose further that you measure position on your paper in centimeters, with the origin being at the origin of your graph. If you plot $\log(x)$ vs $f(x)$, you'll plot points at $(0cm, 1cm), (1cm, 2cm),$ and $(2cm, 3cm)$. If, on the other hand, you use the log paper's log-scale on the x-axis, let's ...

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You are correct. In general, your subtracting for division and fraction for cube root is sound.

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Hint: $6,547.81 = 4,624\cdot \left(1+\frac{0.0042}{12}\right)^{12t}$. Solve for $t$, which is the number of years, then times $12$ to get months.

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