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7

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 $k$th 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} ...


1

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


1

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 $\log_4(\sin^2 A) = -1 = 2\log_4(\sin A)$.


1

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


1

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 ...


1

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


1

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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