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

- anonymous

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- anonymous

logarithms arent fun

- anonymous

I think the theory behind logarithms would lie in number theory

- anonymous

also there is literally only like 4 things you need to any logarithm question

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## More answers

- anonymous

you dont need "a million exercises"

- anonymous

"logarithms arent fun" is that a book or is that what you think?

- anonymous

yeh its a book :p

- anonymous

Im trying to solve a lot of equations, inequalities, doing graphs, etc

- 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

I know the theory behind logarithm in general, but i want something really challenging

- anonymous

Solve \[2\ln(\sqrt{x}) - \ln(x) =1 \]

- anonymous

an example

- anonymous

but its all relatively standard

- anonymous

You want challenging problems, or you want to clear your concept?

- anonymous

want challenging problems

- anonymous

yeh, well logarithms are that challenging to be honest

- anonymous

arent*

- anonymous

You can try one thing. You email me, and I will send you challenging problems

- anonymous

As many as you want

- anonymous

sure. can you give me your e-mail?

- anonymous

About the problem above, elecengineer, I got 0=1. Is that right?

- anonymous

Thanks "pi" I just sent you an email. Can you check it out

- anonymous

It must take a little bit. Im from Ecuador and I suppose your in Europe lol

- anonymous

I haven't yet received your email........... Still waiting

- anonymous

Can you check it out again. Can take a little bit

- anonymous

differentiate \[y= \ln [\frac{ (x-3)^4 \sqrt{x} }{x+1} ]\]

- anonymous

thats prob the hardest question on logarithms every, but its still really easy

- anonymous

Okay code is VILLAMAGUA CONZA LUIS MIGUEL VILLAMAGUA CONZA LUIS MIGUEL

- anonymous

yean. I got 4LN(x-3)+1/2(LN(x))-LN(X+)
cannot remember how to differenciate logs lol

- anonymous

just joking

- anonymous

yeah it is me PI, thanks for your help

- anonymous

I have sent you a problem, but remember, you are not supposed to use a calculator

- anonymous

lols, what was it

- anonymous

By the way, are you a staff of UTPL ?

- anonymous

what was the question!

- anonymous

yeah. From UTPL

- anonymous

You teach there?

- anonymous

No teaching, but a student. Well finishing my major

- anonymous

solve simultaneously \[5^{x+y} = \frac{1}{5}\
and \[5^{3x+2y} =1\]

- anonymous

\[5^{x+y} = \frac{1}{5} , 5^{3x+2y} =1 \]

- anonymous

I just replied you first question PI. Is that all right?

- 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

I dont know much about inequalities with logs.

- anonymous

yeh , they come up a tiny bit in some financial maths topics

- 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

best seen by an example

- 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

solve (1/3)^n > 0.5

- anonymous

solve that, and see what happens

- anonymous

ok

- anonymous

I will put money on you being wrong :P

- anonymous

it looks fairly simple, but there is a trick in it that would catch alot of people

- anonymous

please dont go. Im a little bit slow. I like this :)

- anonymous

well ... have you done the question yet
inequalities with logs behave the exact same as equations

- amistre64

the bigger n is.... the .......

- anonymous

This is what i got

##### 1 Attachment

- 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

What is the other wat? Im excited lol

- anonymous

what is the other way? sorry

- 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

now \[\log_{a} x^r = r \log_{a} x \]

- anonymous

yeah. I know where you go

- anonymous

your bright man

- anonymous

so then you get n ln(1/3) >0.5

- anonymous

*so then you get n ln(1/3) > ln(0.5)

- anonymous

\[n > \frac{\ln(0.5)}{\ln(\frac{1}{3} ) }\]

- anonymous

there the problem people make!

- 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

Yeah I would have fallen there too. Can you email me some of this type of exercices.

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