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12

Yes, more or less, if you meant the following: $$\begin{align} \ln(x+3) + \ln(x - 4) = 0 & \iff \ln((x+3)(x-4)) = 0 \\ \\ &\iff (x+3)(x-4) = 1,\text{ and }x>4\\ \\ & \iff x^2 - x - 13 = 0, \text{ and }x\gt 4 \\ \\ &\iff\cdots\end{align}$$


8

You may write $$\begin{align*} {\large\int}_0^1\frac{\ln^2\ln\left(\frac1x\right)}{1+x+x^2}dx&={\large\int}_0^\infty \left(1-e^{-t}\right)\frac{\ln^2t}{1-e^{-3t}}e^{-t}dt\\ &=\sum_{n=0}^\infty{\large\int}_0^\infty \left(e^{-t}-e^{-2t}\right)e^{-3nt}\ln^2t\:dt\\ &=\sum_{n=0}^\infty \left({\large\int}_0^\infty ...


6

The work below is only partial result. The integral is re-expressed as an infinite series using the trigamma function. First, note the following trigonometric identity: ...


5

In google calculator, "," is mapped to "." and $log(b,a)$ represents $log_{10}{b.a}$


5

Assuming you are allowed to use that $e>2$, you have that $$e^2>2^2=4$$ and therefore$$2^{(e^2)}>2^4=16$$ thus $$\ln 2^{(e^2)} > \ln 16 > \ln e =1$$ (however, here I used that $\ln$ is a monotone increasing function). Now since the left hand side of the above inequality can be written as $e^2\cdot \ln 2$ you have that $$e^2 \ln 2> 1$$ which ...


3

From the given options, I guess that you want to solve the following for $x$ : $$(\log_3(x))^2-\log_\color{red}{3}(x)=2.$$ Let $t=\log_3(x)$. Then, we have $$t^2-t-2=0\iff (t-2)(t+1)=0\iff t=2,-1\iff x=3^2,3^{-1}.$$


3

Let $a$ be a positive real number. Then $$\lim_{n\to\infty}(a^\frac1n-1)\cdot n=\lim_{n\to\infty}\frac{a^{\frac1n}-1}{\frac1n}=\lim_{n\to\infty} \frac{a^\frac1n\cdot\ln a\cdot\frac{-1}{n^2}}{\frac{-1}{n^2}}=\ln a$$ Your result is a special case of this with $a=5$, and $n=2^{20}$


3

Hint: what is $\sum_n \frac{x^n}{n!}$? Then plug $x=\log\log2$.


3

Depends on the value of $a$. For instance, if $a \geq 1 $, then there is no solution. Why? well, $e^{ax} - x = 0 \iff e^{ax} = x $. Hence, your equation is asking where this two functions intersect. IF $a \geq 1 $, then $$ e^{ax} \geq ax + 1 > ax > x $$ Hence, there is no solution. Now, there is indeed a solution if $a \leq \frac{1}{e} $. To see ...


3

using the law of logarithm we have $\ln((x+3)(x-4))=\ln(1)$ thus we have $(x+3)(x-4)=1$


2

There seem to be no issues with your derivation; when I plug in x=2 I get 0.641 to three decimal places...perhaps you substituted in the wrong value?


2

Notice that $(\log_ax)'=\frac{1}{x\ln a}$. SO $$(\log_4(2x^2+1))\big|_{x=2}=\frac{1}{(2x^2+1)\ln 4} \times 4x\big|_{x=2}=\frac{4}{9\ln 2}$$


2

Change the basis to $e$, $\frac{\ln x}{\ln (a+x)}\le 2 \frac{\ln x}{\ln a}$ Then discuss it based on relationship $x, a$ with 1.


2

Beside John ZHANG's recommendation, you could also notice that $$log_{a+x}(x) = log_a(x^2)$$ is satisfied if $x=1$ and $x=\sqrt{a}-a$


2

Remember that $$e^{\ln a}=a$$ Now the derivative should be trivial.


2

I checked this in Mathematica. It really does seem to converge to Log(5) accurately after many iterations: a = N[Sqrt[5], 30]; Do[ a = Sqrt[a], {n, 1, 10000 - 1}] (a - 1)*2^10000 Output: 1.60943791243410037460075933323 compared to: N[Log[5], 30] 1.60943791243410037460075933323


2

We can write this, with $b = \log_{\sqrt a}(a+1) > 0$ as $$b+\frac1b \ge \sqrt6 \iff b \ge \frac1{\sqrt2}+\sqrt\frac32 = c\iff 2\log(a+1) \ge c \log a \iff (a+1)^2 \ge a^c$$ which is obviously true as $a+1 > a$, and $2 > c$


2

If you're careful enough to handle the rounding errors properly on your domain, you could try using some inequality work with some Taylor series. For $x<0$, $\space e^x < 1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} $ $(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4<\log(x)$ Now plug in $x=-2$ and $x=2$ in their ...


2

It seems I posted my question too early, since I found the answer after 2 minutes. Here it is: According to the previous results, we only need to show that $$ \frac{1}{4} \leq \ln2. $$ But $$\frac{1}{4} = \ln e^{\frac{1}{4}}$$ therefore - using the strict monotinicity of the $\ln$ function - we only need to prove that $$e^{\frac{1}{4}} \leq 2 ...


2

It is from Shaw's play In Good King Charles's Golden Days. MRS BASHAM. Oh, do look where youre going, Mr Newton. Someday youll walk into the river and drown yourself. I thought you were out at the university. NEWTON. Now dont scold, Mrs Basham, dont scold. I forgot to go out. I thought of a way of making a calculation that has been puzzling ...


2

If you write its Taylor's expansion then you have: $\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{k}$. For small values of $x$, the values of $x^2, x^3,...$ are small in comparing with $x$(note that positive numbers which are less than 1, will decrees as we multiply themselves), hence we can ignore the terms with degree larger than $1$, and estimate ...


2

Your answer is wrong because when you differentiate with respect to u, you get: $\frac{\mathrm{d} }{\mathrm{d} u}(\frac{4}{\ln u})=\frac{\frac{-4}{u}}{(\ln u)^2}=\frac{-4}{u(\ln u)^2}=\frac{-4}{(x^2+2)(\ln (x^2+2))^2}$ Applying the chain rule by multiplying with $\frac{\mathrm{du} }{\mathrm{d} x}=2x$, we get the answer of $\frac{-8x}{(x^2+2)(\ln ...


1

Re-write the equation as $$y=4(\ln(x^2+2))^{-1}$$ Then use the chain rule and the power rule. Let $u=x^2+2$, so $u'=2x$. Now you have $$y=4(\ln(u))^{-1}$$ This is just the power rule. Let $v=\ln(u)$. Now you have $$y=4v^{-1}$$ $y'$ is just $$-1(v')4v^{-2}$$ $v'$ is just $$\frac{u'}{u}$$ which is $$\frac{2x}{x^2+2}$$ Put this all together and you have ...


1

. Put $v = \ln u$, and $u = x^2 + 2$. $\dfrac{df}{dv} \left(\dfrac{4}{v}\right) = -\dfrac{4}{v^2}$, $\dfrac{dv}{du} = \dfrac{1}{u}$ $\dfrac{du}{dx} = 2x$. Thus by the Chainrule: $\dfrac{df}{dx} = \dfrac{df}{dv}\cdot \dfrac{dv}{du}\cdot \dfrac{du}{dx} = -\dfrac{4}{v^2}\cdot \dfrac{1}{u}\cdot 2x = \dfrac{-4}{\ln^2 (x^2+2)}\cdot \dfrac{1}{x^2+2}\cdot 2x = ...


1

Another possible closed-form $$I_\mu=\frac{2\cdot{_2F_1}\left(\begin{array}c\tfrac12,2\\\tfrac{\mu+5}{2}\end{array}\middle|\,-1\right)}{(\mu+3)(\mu+1)},$$ for $\Re(\mu)>-1$. I don't know how to simplify it further, but I think it's a good idea for integer $\mu$ values to separate the even and the odd $\mu$ cases, since $$\begin{align} I_0 & = ...


1

Since no one else said it clearly in words. One does not have that $$\ln(1 + x) = x$$ for small $x$. One does, however, have that for small values of $x$, $\ln(1+x)$ can be approximated by $x$. As the other answers have already pointed out, this you see from the Taylor expansion $$ \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n} = x - \frac{1}{2}x^2 + ...


1

Hint: One can prove [1] that $$ f(x) \approx f(a) + f'(a)(x - a).$$ for $x$ near $a$.


1

Take the tangent line at of $f(x) = \ln(1+x)$ in $x = 0$. \begin{align*} f(x) & \approx f(0) + f'(0) (x - 0) \\ & = \ln(1+0) + \left[\frac{d}{dx} \ln(1+x)\right]_{x = 0} (x-0) \\ & = 0 + 1 x \\ & = x \end{align*}


1

Look at the equivalent $\text{exp}(x) = 1 + x$ ($x$ teensy) and compare it to the Taylor series.


1

This is an overkill, but I think it is interesting, anyway. We have that $f(x)=e^x \log(1+x)$ is a convex increasing function over $[0,1]$, since $f'(x)=\frac{e^x}{1+x}+f(x)$. Jensen's inequality hence gives: $$ \forall x\in(0,1],\qquad f(x)> f(0)+f'(0)\, x = x, $$ so $f(1)>1$ gives $e\log 2>1$ and $e^{-1}<\log 2$. The argument can be improved, ...



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