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anonymous
 5 years ago
Does anyone knows a really good book where I can learn about logarithms? I want a lot of theory and a million exercises.
anonymous
 5 years ago
Does anyone knows a really good book where I can learn about logarithms? I want a lot of theory and a million exercises.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the theory behind logarithms would lie in number theory

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also there is literally only like 4 things you need to any logarithm question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you dont need "a million exercises"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"logarithms arent fun" is that a book or is that what you think?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im trying to solve a lot of equations, inequalities, doing graphs, etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im 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
 5 years ago
Best ResponseYou've already chosen the best response.0I know the theory behind logarithm in general, but i want something really challenging

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solve \[2\ln(\sqrt{x})  \ln(x) =1 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but its all relatively standard

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You want challenging problems, or you want to clear your concept?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0want challenging problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeh, well logarithms are that challenging to be honest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can try one thing. You email me, and I will send you challenging problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure. can you give me your email?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0About the problem above, elecengineer, I got 0=1. Is that right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks "pi" I just sent you an email. Can you check it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It must take a little bit. Im from Ecuador and I suppose your in Europe lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I haven't yet received your email........... Still waiting

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you check it out again. Can take a little bit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate \[y= \ln [\frac{ (x3)^4 \sqrt{x} }{x+1} ]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats prob the hardest question on logarithms every, but its still really easy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay code is VILLAMAGUA CONZA LUIS MIGUEL VILLAMAGUA CONZA LUIS MIGUEL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yean. I got 4LN(x3)+1/2(LN(x))LN(X+) cannot remember how to differenciate logs lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah it is me PI, thanks for your help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have sent you a problem, but remember, you are not supposed to use a calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the way, are you a staff of UTPL ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what was the question!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No teaching, but a student. Well finishing my major

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0solve simultaneously \[5^{x+y} = \frac{1}{5}\ and \[5^{3x+2y} =1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[5^{x+y} = \frac{1}{5} , 5^{3x+2y} =1 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just replied you first question PI. Is that all right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there 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
 5 years ago
Best ResponseYou've already chosen the best response.0I dont know much about inequalities with logs.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeh , they come up a tiny bit in some financial maths topics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0inequalities 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
 5 years ago
Best ResponseYou've already chosen the best response.0best seen by an example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When 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
 5 years ago
Best ResponseYou've already chosen the best response.0solve that, and see what happens

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will put money on you being wrong :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it looks fairly simple, but there is a trick in it that would catch alot of people

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please dont go. Im a little bit slow. I like this :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well ... have you done the question yet inequalities with logs behave the exact same as equations

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the bigger n is.... the .......

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0long 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
 5 years ago
Best ResponseYou've already chosen the best response.0What is the other wat? Im excited lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the other way? sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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
 5 years ago
Best ResponseYou've already chosen the best response.0now \[\log_{a} x^r = r \log_{a} x \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. I know where you go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then you get n ln(1/3) >0.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*so then you get n ln(1/3) > ln(0.5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[n > \frac{\ln(0.5)}{\ln(\frac{1}{3} ) }\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there the problem people make!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mm, 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
 5 years ago
Best ResponseYou've already chosen the best response.0ln(a) is negative for all a in the interval 0<a<1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah I would have fallen there too. Can you email me some of this type of exercices.
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