anonymous
  • anonymous
Does anyone knows a really good book where I can learn about logarithms? I want a lot of theory and a million exercises.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
logarithms arent fun
anonymous
  • anonymous
I think the theory behind logarithms would lie in number theory
anonymous
  • anonymous
also there is literally only like 4 things you need to any logarithm question

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anonymous
  • anonymous
you dont need "a million exercises"
anonymous
  • anonymous
"logarithms arent fun" is that a book or is that what you think?
anonymous
  • anonymous
yeh its a book :p
anonymous
  • anonymous
Im trying to solve a lot of equations, inequalities, doing graphs, etc
anonymous
  • anonymous
im out trying to promote my online tutoring....i tutor math and physics from low - hgh level....if u ever need it check me out http://mathphystut.tripod.com
anonymous
  • anonymous
I know the theory behind logarithm in general, but i want something really challenging
anonymous
  • anonymous
Solve \[2\ln(\sqrt{x}) - \ln(x) =1 \]
anonymous
  • anonymous
an example
anonymous
  • anonymous
but its all relatively standard
anonymous
  • anonymous
You want challenging problems, or you want to clear your concept?
anonymous
  • anonymous
want challenging problems
anonymous
  • anonymous
yeh, well logarithms are that challenging to be honest
anonymous
  • anonymous
arent*
anonymous
  • anonymous
You can try one thing. You email me, and I will send you challenging problems
anonymous
  • anonymous
As many as you want
anonymous
  • anonymous
sure. can you give me your e-mail?
anonymous
  • anonymous
About the problem above, elecengineer, I got 0=1. Is that right?
anonymous
  • anonymous
Thanks "pi" I just sent you an email. Can you check it out
anonymous
  • anonymous
It must take a little bit. Im from Ecuador and I suppose your in Europe lol
anonymous
  • anonymous
I haven't yet received your email........... Still waiting
anonymous
  • anonymous
Can you check it out again. Can take a little bit
anonymous
  • anonymous
differentiate \[y= \ln [\frac{ (x-3)^4 \sqrt{x} }{x+1} ]\]
anonymous
  • anonymous
thats prob the hardest question on logarithms every, but its still really easy
anonymous
  • anonymous
Okay code is VILLAMAGUA CONZA LUIS MIGUEL VILLAMAGUA CONZA LUIS MIGUEL
anonymous
  • anonymous
yean. I got 4LN(x-3)+1/2(LN(x))-LN(X+) cannot remember how to differenciate logs lol
anonymous
  • anonymous
just joking
anonymous
  • anonymous
yeah it is me PI, thanks for your help
anonymous
  • anonymous
I have sent you a problem, but remember, you are not supposed to use a calculator
anonymous
  • anonymous
lols, what was it
anonymous
  • anonymous
By the way, are you a staff of UTPL ?
anonymous
  • anonymous
what was the question!
anonymous
  • anonymous
yeah. From UTPL
anonymous
  • anonymous
You teach there?
anonymous
  • anonymous
No teaching, but a student. Well finishing my major
anonymous
  • anonymous
solve simultaneously \[5^{x+y} = \frac{1}{5}\ and \[5^{3x+2y} =1\]
anonymous
  • anonymous
\[5^{x+y} = \frac{1}{5} , 5^{3x+2y} =1 \]
anonymous
  • anonymous
I just replied you first question PI. Is that all right?
anonymous
  • anonymous
there fairly standard lol all of logarithms are fairly standard. I have no idea why people find them hard , all it is is 4formulas to remember
anonymous
  • anonymous
I dont know much about inequalities with logs.
anonymous
  • anonymous
yeh , they come up a tiny bit in some financial maths topics
anonymous
  • anonymous
inequalities and logs dont come up alot there is however one thing which you do need to look out for when solving inequalities with logs
anonymous
  • anonymous
best seen by an example
anonymous
  • anonymous
When I was in high school logarithms seems to be so hard. Now I tried them again and they look so easy. Im confused. I thought I would never be done with logs
anonymous
  • anonymous
solve (1/3)^n > 0.5
anonymous
  • anonymous
solve that, and see what happens
anonymous
  • anonymous
ok
anonymous
  • anonymous
I will put money on you being wrong :P
anonymous
  • anonymous
it looks fairly simple, but there is a trick in it that would catch alot of people
anonymous
  • anonymous
please dont go. Im a little bit slow. I like this :)
anonymous
  • anonymous
well ... have you done the question yet inequalities with logs behave the exact same as equations
amistre64
  • amistre64
the bigger n is.... the .......
anonymous
  • anonymous
This is what i got
1 Attachment
anonymous
  • anonymous
long way to go about things also you would need to use change of basic at the end if you actually wanted to get a number for it
anonymous
  • anonymous
What is the other wat? Im excited lol
anonymous
  • anonymous
what is the other way? sorry
anonymous
  • anonymous
well the way I was taught was to take logarithms of both sides ( it doesnt matter what base, as long as you use the same on both sides ) so I will take ln of both sides so \[\ln [( \frac{1}{3})^n] = \ln(0.5) \]
anonymous
  • anonymous
now \[\log_{a} x^r = r \log_{a} x \]
anonymous
  • anonymous
yeah. I know where you go
anonymous
  • anonymous
your bright man
anonymous
  • anonymous
so then you get n ln(1/3) >0.5
anonymous
  • anonymous
*so then you get n ln(1/3) > ln(0.5)
anonymous
  • anonymous
\[n > \frac{\ln(0.5)}{\ln(\frac{1}{3} ) }\]
anonymous
  • anonymous
there the problem people make!
anonymous
  • anonymous
mm, I divide both sides by ln(1/3) , but am I allowed to do that? well , ln(1/3) is actually a negative number , so I must flip the inequality sign
anonymous
  • anonymous
ln(a) is negative for all a in the interval 0
anonymous
  • anonymous
Yeah I would have fallen there too. Can you email me some of this type of exercices.

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