# Tag Info

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 $$\ln 3 (\ln 3 + \ln x)=\ln 4 ( \ln 4 + \ln x)$$ Thus $$\ln x =\frac{(\ln 3)^2 - (\ln 4)^2}{\ln 4 - \ln 3}=-(\ln 4+\ln 3)=-\ln 12=\ln \frac{1}{12}.$$ 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 inequalitye^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}$$can you proceed? 2 You approach is good, but there is another which is worthy of consideration. Define h : (0,\infty) \rightarrow \mathbb{R} by $$h(x) = x \text{ln}(x) - x + 1.$$ and show that this function is non-negative if and only if f(x) \ge g(x). Then subject h to a standard functional analysis in order to determine its range. Compute ... 2 If you allow, you can solve for x through Lagrange Inversion Theorem.$$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)\ne0f^{-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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