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13

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 ...


8

If the textbook is making clear that you should read $\log n$ to mean $\ln n$, then yes, barring any subscript for a base other than $e$, $n \log(n) = n\ln(n)$.


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

If $\log = \ln$, then yes, indeed, $n\log n = n\ln n$. Note, however, that this is utterly unimportant if you are looking at it from the case of the big $O$ notation, which is highly likely, since $$n\log n \in O(n\ln n)$$ and $$n\ln n \in O(n\log n)$$ or, in other words, $$O(n\log n)=O(n\ln n)$$


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 ...


4

It is one of the logarithmic identities: $$\log(a^b) = b\log(a)$$


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

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 ...


3

Normally, the base of a logarithm must be specified as $\log_a$. A very common convention is $\ln \equiv \log_e$. However, $log$ without subscript can mean a few different things based on the context (and therefore must be always explicitly stated). $log \equiv log_{10}$ is very common in many mathematical books and publications. $log \equiv \log_2$ in ...


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

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 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

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 ...


2

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


1

$v=\ln x^2=u^2$ ?? it would be right if $(ln x)^2$ it should be $v=\ln x^2=2u$


1

Even if you don't know what $\sum\frac{x^n}{n!}$ is, the first two terms give $1+\log\log2\approx0.63\ldots>\frac35$. Since this is an alternating series and $1+\log\log2$ is a partial sum ending on a negative term, whatever the sum is it will be greater than that $0.63\ldots$.


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

If it were to be asked in the exam, you could save some time in the simplification exploiting the fact that $a^3-b^3 = (a-b).(a^2+ab+b^2)$ Rearanging in the following way $$\frac{1}{6}\left((x+3)^3 - (x+2)^3 - \left((x+1)^3 - x^3)\right)\right)$$ $$\frac{1}{6}\left((x+3)^2 + (x+2)^2+(x+3)(x+2) - \left((x+1)^2 + x^2+x(x+1))\right)\right)$$ ...


1

A good choice of substitution might help cut through the clutter of logs and nested parentheses, giving a visually pleasing and symmetrical solution. Let $\text{L}^mn=(\log_2 n)^m$ and put $u=L3=\log_23$. Hence $$\begin{align} \text{L}^3 3&=(L3)^3&=u^3\\ \text{L}^3 6&=(L3+L2)^3&=(u+1)^3\\ \text{L}^3 12&=(L2+L4)^3&=(u+2)^3\\ ...


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, ...


1

Yet another argument, using the inequalities $2<e<4$: From $e<4$ we get $1<2\ln2$ so $\ln2>1/2$; from $2<e$ we get $1/2>1/e$. Conclusion: $\ln2>e^{-1}$.


1

General solution to the equation $|x-1|^{log(x)^2-2log(x)}=|x-1|^3$: $log(x)^2-2log(x)=3$ $|x-1|=1$ $[|x-1|=0]\wedge[log(x)^2-2log(x)\neq0]$ Pitfalls to watch for: Make sure that the solution doesn't yield $0^0$ Make sure that the solution doesn't yield $log(0)$


1

It surely isn't a nice solution, but you could proceed as follows: $$ log_{√a}(a+1)+log_{a+1}\sqrt a=\frac{\ln(a+1)}{\ln(\sqrt a)}+\frac{\ln(\sqrt a)}{\ln(a+1)}=\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)} $$ What we now need to prove is: $$ \frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}≥\sqrt6\iff \left( ...



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