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8

Consider function $$f(x)=x - \frac{1000}{\log 2} \log (x)$$ $$f'(x)=1-\frac{1000}{x \log (2)}$$ The function goes through a minimum (by second derivative test) at $x=\frac{1000}{\log (2)}$. So, let us start Newton method which will generate the following iterates $$x_0=2000$$ $$x_1=34175.5$$ $$x_2=14218.2$$ $$x_3=13747.7$$ $$x_4=13746.8$$ which is the ...


8

Solve $$x-\frac{1000}{\log(2)}\log(x)=0$$ for $x$. Substitute $x=e^t$: $$e^t-\frac{1000}{\log(2)}t=0;$$ subtract $e^t$ from both sides: $$-\frac{1000}{\log(2)}t=-e^t;$$ multiply both sides by $\dfrac{\log(2)}{1000}$: $$-t=-\frac{\log(2)}{1000}e^t;$$ divide both sides by $e^t$: $$-t\frac{1}{e^t}=-\frac{\log(2)}{1000};$$ rewrite $1/e^t=e^{-t}$: ...


8

Take the natural logarithms of both sides, then \begin{equation} \ln 3 (\ln 3 + \ln x)=\ln 4 ( \ln 4 + \ln x) \end{equation} Thus \begin{equation} \ln x =\frac{(\ln 3)^2 - (\ln 4)^2}{\ln 4 - \ln 3}=-(\ln 4+\ln 3)=-\ln 12=\ln \frac{1}{12}. \end{equation} Since $\ln x$ is injective, $x=\frac{1}{12}$.


5

There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$. $$ \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx ...


4

I thought it might be instructive to present a way forward that does not rely on calculus, but rather elementary analysis only. In THIS ANSWER and THIS ONE, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality $$e^x\ge 1+x \tag 1$$ Setting $x=-z/(z+1)$ into ...


4

For all $x \in \mathbb{R}$, $$e^x \geq 1 + x$$ Taking log on both sides we get, $$\ln (1 + x) \leq x, \forall x > -1$$ Substituting $x = \frac{1}{k}, k \notin [0, -1]$, we get, $$\displaystyle{\ln \left(1 + \frac{1}{k}\right) \leq \frac{1}{k}}$$ Substituting $x = \frac{-1}{k + 1}, k \notin [0, -1]$, we get, $$\ln \left(\frac{k}{k + 1}\right) \leq ...


4

Note that $\ln(a^x)=x\ln(a)$ Thus $$ln(e^{2x}) = ln(4/3)$$ $$(2x)ln(e) = ln(4/3)$$ $$2x = ln(4/3)$$ since $\ln(e)=1$ $$x=\frac{1}{2}\ln(\frac{4}{3})$$


4

Recall the logarithm property $$\frac{\ln x^2}{\ln x} = \frac{2\ln x}{\ln x} = 2.$$ But this is only true when $x>0$ and $x\neq1$. Otherwise, there is a "hole" there; a removable discontinuity. Notice that this is difficult not to graph, so graphing tools usually just fill the hole/graph over it.


3

$$\frac{\ln(x^{2})}{\ln(x)} = \frac{2\ln(x)}{\ln(x)} = 2 $$


3

Some hints: $$\log_c(A\cdot B) = \log_c(A) + \log_c(B)$$ $$\log_c(D^n) = n\log_c(D)$$ Then, if $A=3$, $B = f(n)^n$ ... what can be made of $$\log_2\left(3f(n)^n\right)$$


3

$$x-\frac{1000\ln(x)}{\ln(2)}=0\Longleftrightarrow$$ $$-\frac{1000\ln(x)}{\ln(2)}=-x\Longleftrightarrow$$ $$\frac{1000\ln(x)}{\ln(2)}=x\Longleftrightarrow$$ $$1000\ln(x)=x\ln(2)\Longleftrightarrow$$ $$e^{1000\ln(x)}=e^{\ln(2)x}\Longleftrightarrow$$ $$x^{1000}=2^x\Longleftrightarrow$$ ...


2

First of all, I assume that you mean the logarithm to the base $2$, i.e. that your problem is to calculate: $$\log_2(\sqrt[4]{4}).$$ Let's recall a rule for taking roots: Suppose $x>0$, and that $n$ is an integer. Then it is true that $$\sqrt[n]{x} = x^{1/n}.$$ I want to use this in your problem for the expression inside the logarithm. We get that ...


2

The answer should be $6.25$. \begin{align} & 4 \sinh (2 \ln 2) - \cosh(\ln2 ) \\ =& 2 \left(e^{2\ln2}-e^{-2\ln2} \right) - \frac{e^{\ln2}+e^{-\ln2}}{2}\\ =& 2(4-0.25)-\frac{2+0.5}{2}\\ =& 7.5-1.25 = 6.25. \end{align}


2

As already said in comments, there are quite many things you can do if you consider the function and its derivatives $$f(x)=x+3-3^x$$ $$f'(x)=1-3^x \log (3)$$ $$f''(x)=-3^x \log ^2(3)$$ The first derivative cancels for $$x_*=-\frac{\log (\log (3))}{\log (3)}\approx -0.085606$$ For this value $$f(x_*)=3-\frac{1}{\log (3)}-\frac{\log (\log (3))}{\log ...


2

HINT: set $$f(x)=\ln(x)-\frac{x-1}{x}$$ then we get $$\lim_{x \to 0+}f(x)=+\infty$$ further is $$f'(x)=\frac{x-1}{x^2}$$ and $$f''(x)=-\frac{x-2}{x^3}$$ can you proceed?


2

You approach is good, but there is another which is worthy of consideration. Define $h : (0,\infty) \rightarrow \mathbb{R}$ by \begin{equation} h(x) = x \text{ln}(x) - x + 1. \end{equation} and show that this function is non-negative if and only if $f(x) \ge g(x)$. Then subject $h$ to a standard functional analysis in order to determine its range. Compute ...


2

If you allow, you can solve for $x$ through Lagrange Inversion Theorem. $$y=x^{x+1}$$ Invert it... I now have time to work the problem out. $$f(x)=x^{x+1}$$ $$f(1)=1,f'(1)\ne0$$ $$f^{-1}(x)=1+\sum_{n=1}^{\infty}\lim_{w\to1}\frac{(x-1)^n}{n!}\frac{d^{n-1}}{dw^{n-1}}\left(\frac{w-1}{w^{w+1}-1}\right)^n$$ This is not reducible as far as I know, so you ...


2

Set $f:[k,k+1]\to\mathbb{R}$ given by $f(x)=\log(x)$ Then, $f$ is continuous in $[k,k+1]$ and differentiable in $(k,k+1)$. Thus, there is $\xi\in(k,k+1)$ such that $\frac{f(k+1)-f(k)}{k+1-k}=f^\prime(\xi)$. That is $\log(k+1)-\log(x)=\frac{1}{\xi}$ for some $\xi\in(k,k+1)$. Then ...


2

Hint: $a^{\log_{a}(x)} = x$, so: $$x^2+2ax = 4x-4a-13$$ $$\Rightarrow x^2+2(a-2)x+(4a+13) = 0$$ And use the quadratic formula.


2

$$(3x)^{\ln(3)}=(4x)^{\ln(4)} \quad \iff\quad \frac{3^{\ln(3)}}{4^{\ln(4)}}=x^{\ln(4)-\ln(3)}\quad \iff\quad x = \bigg(\frac{3^{\ln(3)}}{4^{\ln(4)}}\bigg)^{\frac1{\ln(4)-\ln(3)}}$$ If you want a "nicer" solution. As pointed out in the comments, take the logarithm: We have ...


2

This doesn't require using logarithms to get an answer. $$(3x)^{\ln 3}=(4x)^{\ln 4}$$ $$3^{\ln 3}x^{\ln 3} = 4^{\ln 4}x^{\ln 4}$$ $$\frac{x^{\ln 4}}{x^{\ln 3}} = \frac{3^{\ln 3}}{4^{\ln 4}}$$ $$x^{\ln 4 - \ln 3} = \frac{3^{\ln 3}}{4^{\ln 4}}$$ $$x^{\ln \frac43} = \frac{3^{\ln 3}}{4^{\ln 4}}$$ $$x = \left(\frac{3^{\ln 3}}{4^{\ln 4}}\right)^{\frac{1}{\ln ...


2

Here, $\lg^2 n$ stands for $(\log_2 n)^2$. Your question is then to study the asymptotic behavior of $\sum_{k=0}^{\log_2 n} \log_2 \frac{n}{2^k}$.${}^{(\dagger)}$ Below are two different methods, one using knowledge of $\sum_{\ell=1}^m \ell = \frac{m(m+1)}{2}$ and yielding a sharp estimate; the second requiring no prior knowledge, but only giving a ...


2

I find exponentials easier to deal with than logs. The first log equation says that $2=6^a$, and the second says that $3=5^b$. Rewrite $2=6^a$ as $2=2^a\cdot 3^a$, and then as $2=2^a(5^b)^a$. We obtain the equation $$2=2^a\cdot 5^{ab}.$$ This can be rewritten as $$2^{1-a}=5^{ab}.$$ Taking logs to the base $5$,we get $(1-a)\log_5 2=ab$. Since $a\ne 1$, the ...


2

An alternative, although I agree with the exponential approach, too! We're given: $\log_6 2 = a$ and $\log_5 3 = b$ We want: $\log_5 2$ We must recall our logarithms rules. There are too many bases happening here, so let's fix that! The change of base formula gives us $\log_6 2 = \frac{\log_5 2}{\log_5 6}$ -- I thought to do this because we're looking ...


2

Notice, a few things: $$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$ $$\ln(e)=\log_e(e)=\frac{\ln(e)}{\ln(e)}=1$$ $$\ln(x)=\log_e(x)=\frac{\ln(x)}{\ln(e)}=\frac{\ln(x)}{1}=\ln(x)$$ $$\ln(a^x)=x\ln(a)\space\space\space\text{when}\space a,x\space\text{are positive}$$ $$\ln\left(\frac{a}{x}\right)=\ln(a)-\ln(x)\space\space\space\text{when}\space a,x\space\text{are ...


2

Yes, this is true. This is equivalent to proving that, for any $a > 0$, we have $$ \frac{\ln x}{x^a} \xrightarrow[x\to\infty]{}0 $$ (you can see it by setting $a=\frac{1}{m}$ from your question).\; which itself is equivalent to showing $$ \frac{a\ln x}{x^a} = \frac{\ln x^a}{x^a} \xrightarrow[x\to\infty]{}0 $$ so, at the end of the day, it is sufficient ...


2

No calculus required.Taking logs to any base $b=1+r$ with $r>0,$ then for $x>(1+r)^2$ we have $\log_b x>2.$ So for $x>(1+r)^2$ let $n_x$ be the positive integer such that $$n_x\leq \log_bx<n_x+1.$$ We have then $b^{n_x}\leq x<b^{n_x+1}$. And since $n_x\geq 2$ and $r>0$, we have, by the Binomial Theorem, $$x\geq b^{n_x}=(1+r)^{n_x}=1+r ...


2

It is clear that any solution must have $3x + 1 > 0$, or $x > -1/3$. For $x > -1/3$, the function $f(x) = 8^x (3x + 1)$ is increasing, hence the equation $f(x) = 4$ has at most one solution. But $x = 1/3$ is a solution, so it is the only one.


1

You ended up with the system $\log (1+a)=2-a$ where $a=3x$. Observe that $2-a$ is strictly decreasing and $\log (1+a)$ is strictly increasing function of $a$. So there is a unique solution thanks to intermediate value theorem.


1

You should look at the properties of log. They can be found all over the web (...and in many books too).



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