Question

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

Answer 1

logarithms arent fun

Answer 2

I think the theory behind logarithms would lie in number theory

Answer 3

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

Answer 4

you dont need "a million exercises"

Answer 5

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

Answer 6

yeh its a book :p

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

an example

Answer 12

but its all relatively standard

Answer 13

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

Answer 14

want challenging problems

Answer 15

yeh, well logarithms are that challenging to be honest

Answer 16

arent*

Answer 17

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

Answer 18

As many as you want

Answer 19

sure. can you give me your e-mail?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

Okay code is VILLAMAGUA CONZA LUIS MIGUEL VILLAMAGUA CONZA LUIS MIGUEL

Answer 28

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

Answer 29

just joking

Answer 30

yeah it is me PI, thanks for your help

Answer 31

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

Answer 32

lols, what was it

Answer 33

By the way, are you a staff of UTPL ?

Answer 34

what was the question!

Answer 35

yeah. From UTPL

Answer 36

You teach there?

Answer 37

No teaching, but a student. Well finishing my major

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

I dont know much about inequalities with logs.

Answer 43

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

Answer 44

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

Answer 45

best seen by an example

Answer 46

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

Answer 47

solve (1/3)^n > 0.5

Answer 48

solve that, and see what happens

Answer 49

ok

Answer 50

I will put money on you being wrong :P

Answer 51

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

Answer 52

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

Answer 53

well ... have you done the question yet


inequalities with logs behave the exact same as equations

Answer 54

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

Answer 55

This is what i got

Answer 56

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

Answer 57

What is the other wat? Im excited lol

Answer 58

what is the other way? sorry

Answer 59

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

Answer 60

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

Answer 61

yeah. I know where you go

Answer 62

your bright man

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

there the problem people make!

Answer 67

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

Answer 68

ln(a) is negative for all a in the interval 0<a<1

Answer 69

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