I understand how both graphs are drawn, but I do not understand why you cannot just convert one into another. It feels natural to me to just convert $2\log x$ to $\log x^2$ and not have to worry about the domain restriction.
123333
These two are supposed to be equivalent because of the properties of logarithms, but the domains of $\ln(x^2)$ and $2\ln x$ seem different to me. For example, if I substitute $x=-1$ into the first, I get 0. But in the second, I get a non-real answer.   Why is this? The domains of these functions, when graphed, seem different as well. But everything I've been taught so far, and most of the things I can find on the web, do not explain this inconsistency. Typically, I consider the property $\ln(a^b) = b \ln a$ to be true...  I am only a senior in high school, currently in Calculus I, but hopefully the explanation won't be too outside of the realm of my understanding. But even if it is, I would still like to know the answer.   Thanks in advance
55399
I'm currently doing a couple of exercises on logarithmic expressions, and I was a bit confused when presented with the following: $5^{\log_5 17}$.  The answer is $17$, which is the argument of the logarithm in the exponent, but I don't understand the reason why.  A previous question in the exercise was $12^{\log_{12} 144}$, which is pretty straight forward since $\log_{12} 144=2$ and $12^2=144$ but since $\log_5 17$ is an irrational number, I couldn't calculate it the same way.  So my question is, why is $x^{\log_x n}=n$?  [Update]  I've now managed to reason this out and realized how simple it is.  Basically, if $y=\log_x n$, then $y$ is the number that  $x$ must be raised to to become $n$; therefore, if I raise $x$ to that power $y$, I will naturally get $n$.  I guess my problem was that I tried to work out the question mechanically instead of intelligently and that's why I failed, because once I realized that I can't calculate $\log_5 17$, I was stumped.  Now that I reasoned it out, as explained in the previous paragraph, I understand it.
42073
is there a general solution to that? I have seen simple examples for Y+X=Z but I was wondering how this would be with rescaling?
7100
$f(n) = n + (\log n)^{2} , g(n) = n + \log(n^{2} ).$  Now so far I've done $g(n) = n+ 2\log(n)$  and then I think since its the same change to both, I can remove $n$.  leaving me with $f(n) = (\log n)^{2}$ and $g(n) = 2\log(n)$ now if I do $f(n)/g(n)$ I  get  $((\log n)^{2}) / (2\log n)$   and since the numerator is growing exponentially and the bottom is growing linearly? As we approach positive infinity the function approaches positive infinity. Thus $f(n)$ is $\omega$ of $g(n)$?  I don't know if this proof is done right, if its what is a better alternative? I am still new to this and learning!
92865
I am trying to find the intersection points of $2^x$ and $x^2$.  To do this, I set the two functions equal to each other: $2^x = x^2$. It seems intuitive to take the log base two of both sides of the equations, and I got: $x = \log_2{x^2}$.  I'm not entirely sure how to proceed from here. Can someone please help me?
98951
There are so many available bases. Why is the strange number $e$ preferred over all else?  Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
5503
I'm currently trying to prove the following:    And this is what I've come up with so far:    And now I'm stuck. Perhaps the next step might be to eliminate for all x?  I'm sorry for using links, but it seems like I'm not allowed to do it otherwise due to needed reputation point.  Help would be appreciated.
83091
Why is it such a good idea to plot the logarithm of the form factor vs $Q^2$ in . It seems arbitrary to me.
77293
I realize that I lack any numerical intuition for logarithms. For example, when comparing two logarithms like $\log_78$ and $\log_89$, I have to use the change-of-base formula and calculate the values with a calculator in order to determine which of the two is larger.  Can anyone demonstrate an algebraic/analytic method to find out which of the two is larger without using a calculator?
86263
I'm trying to graph the function $y = \frac{x^2-4}{x-2}$.  I factored the numerator into $(x+2)(x-2)$, and then canceled the common factor (x-2), leaving the function as $y = x + 2$.    Is it acceptable to graph the original function as $y = x + 2$, a straight line with domain and range all real numbers?  Or do you have to restrict the domain as in the original function?  Thanks, Sean
120469
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ? i know that its "1", but why? How can i calculate this?  Thank you very very much =)
49360
I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log: $$x\ln \left( 2\right) =2\ln \left( x\right)$$ $$\dfrac {\ln \left( 2\right) }{2}=\dfrac {\ln \left( x\right) }{x}$$ I don't know what to do from here so I decided to try another method: $$2^{x}=2^{\log _{2}\left( x^{2}\right) }$$ $$x=\log _{2}\left( x^{2}\right)$$ And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple&hellip;
71361
$\lim\limits_{(x,y) \to (0,0)}\ {xy\log(x^2+y^2)}$  I tried this:  $$xy\log(x^2+y^2)=xy\log(x^2)+xy\log(1+y^2/x^2)$$  $|xy\log(x^2)|=2|x||y||\log|x||$ (because I use $(x,y)\neq(0,0)$) $\le 2|x||y|||x|-1|$ and taking the limit, $\lim_{(x,y) \to (0,0)}\ {xy\log(x^2)}=0$.  Since the other member has two variables in the argument of $\log$, I don't know how to continue. I know the limit of this function is zero because proceeding with other methods I've seen the limit is zero.
13721
$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$  Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?
46957
In  in these notes, it is said that  Since $x − a$ has leading coefficient $1$, which is a unit, we may use the Division Algorithm...  Why is this true? I thought that the Division Algorithm is only guaranteed to work in Euclidean domains not any integral domain.
64071
I've already read  question about the difference between discrete and non discrete logarithms. But I still have problems to clarify my mind about why the Discrete Logarithm Problem is computationally hard while the NON Discrete Logarithm Problem is not, as I read in my Cryptography and Security book.  What are the computation differences? Why can I, for example, easily break a non discrete logarithm cipher but not a discrete one?
108993
It is easy to show $\log$ isn't a polynomial (no continuous extension to $\mathbb{R}$). More challenging is showing it isn't rational.   Suppose it were a rational function. Then write, the fraction in lowest terms  $$\log x =\frac{G(x)}{Q(x)} \Longleftrightarrow\frac{G(x)}{\log x} = Q(x)$$  Clearly, as $x \to 0$, $Q(x) \to 0$. However, $Q(x)$ then has $x$ as factor so that   $$\frac{G(x)}{x\log x} = Q_2(x)$$  It is well known that $x \log x \to 0$ as $x \to 0$, so for $Q_2(x)$ to have a finite limit as $x \to 0$, which it must since it is a polynomial, $G(x) \to 0$ as $x \to 0$ so that $x$ is a factor of $G(x)$. This contradicts the assumption that $\frac{G}{Q}$ was in lowest terms.  If there is something wrong with this, please comment, but my main question is      What are some other ways to prove that $\log x$ isn't a rational function?