law of logarithms

[Vastor]

i having problem of understanding log, the question is

 

solve the equation [math] log_2 2p - log_2 (1-3p) = 1[/math]

 

so, i calculate using this way

 

[math] log_2 2p - log_2 (1-3p) = 1[/math]

[math] log_2 2p - log_2 (1-3p) = log_2 2[/math]

[math] log_2 \frac{2p}{(1-3p)} = log_2 2[/math]

and I'm consider [math]\frac{2p}{(1-3p)} = 2[/math]

so [math] p = \frac{1}{4}[/math]

confirmed this?

Edited May 17, 2011 by Vastor

[DJBruce]

I am not really sure what you were trying to do, because your work is not correct the second line is not really needed to solve this equation.

As for solving this why not try so of the properties of logs. For example,

 

[math]log(a)-log(b)=log(\frac{a}{b})[/math]

 

might be useful since you could then combine the entire left side of the equation into one log meaning you could "delogify" ie: take the inverse of both sides to end up with a fairly easy equation to solve.

 

This is to much like homework for me to do any more, but if you post you work, or have any question feel free to ask.

 

Yep, thats right, but your second step is really not needed.

Edited May 17, 2011 by DJBruce

[Vastor]

when i checked

 

[math] log_2 2\frac{1}{4} - log_2 (1-3\frac{1}{4}) = log_2 2[/math]

 

[math] log_2 \frac{1}{2} - log_2 \frac{1}{4} = log_2 2[/math]

 

[math] log_2 \frac{1}{4} = log_2 2[/math] and yes, this is fail, i miss something somewhere?

 

from my understanding, the question of 'solve the equation' would mean to be find the value of the unknown 'p', doesn't ?

[imatfaal]

You sure DJB - looks fine to me.

[Vastor]

and there are a law of logs [math] log_a a = 1[/math]

[DJBruce]

My bad, wasn't quite thinking there.

[imatfaal]

log_x(x) = 1

 

 

I would put a bigger bracket around it to be sure - but I think you are sound

 

 

[math] log_2 \left( \frac{2p}{(1-3p)} \right) = log_2 2[/math]

 

when i checked

 

[math] log_2 2\frac{1}{4} - log_2 (1-3\frac{1}{4}) = log_2 2[/math]

 

[math] log_2 \frac{1}{2} - log_2 \frac{1}{4} = log_2 2[/math]

 

[math] log_2 \frac{1}{4} = log_2 2[/math] and yes, this is fail, i miss something somewhere?

 

from my understanding, the question of 'solve the equation' would mean to be find the value of the unknown 'p', doesn't ?

 

The reason you got this bit wrong is that

 

[math] log_2 \frac{1}{2} - log_2 \frac{1}{4} \not= log_2 \frac{1}{4}[/math]

 

 

If you think about it log₂ (1/2) has to be -1 as 2^-1 = 1/2 and similarly log₂(1/4) has to be -2 as 2^-2=1/4

 

and -1-(-2) = 1 = log₂(2)

[Vastor]

log_x(x) = 1

 

 

I would put a bigger bracket around it to be sure - but I think you are sound

 

 

[math] log_2 \left( \frac{2p}{(1-3p)} \right) = log_2 2[/math]

 

 

 

The reason you got this bit wrong is that

 

[math] log_2 \frac{1}{2} - log_2 \frac{1}{4} \not= log_2 \frac{1}{4}[/math]

 

 

If you think about it log₂ (1/2) has to be -1 as 2^-1 = 1/2 and similarly log₂(1/4) has to be -2 as 2^-2=1/4

 

and -1-(-2) = 1 = log₂(2)

 

 

gotcha, ha3, my bad

 

 

[Vastor]

ermm, another problem...

 

what i'm gonna do if [math] log_3 9 * log_4 5 [/math]

 

i'm not learn anything about 'multiply' or 'dividing' logarithms yet, but it's appear to my exam just now

[imatfaal]

Two things - Firstly in general, you need to learn how to convert log base x into log base y, cos dealing with separate bases is a real pain. I would convert both to natural logs. Secondly, in particular [math] log_3(9) = ? [/math]

 

I think any multiplication or division you come across will be looking for a re-arrangement into something more manageable rather directly tackling the problem

[Vastor]

i think i should move to the 'real question', because [math] log_3(9) [/math] is just my creation. So it's said

 

Evaluate [math] log_2 9 * log_3 4 * log_4 8 [/math] 2marks

 

 

if I'm not wrong, converting log base would be using this formula

 

[math] log_a(b) = \frac {log_c b}{log_c a} [/math]

 

so [math] log_2(9) = \frac {log_10(9)}{log_10(2)} [/math] * [math] log_3(4) = \frac {log_10(4)}{log_10(3)} [/math] * [math] log_4(8) = \frac {log_10(8)}{log_10(4)} [/math]

 

don't know how to 'zoom out' the '0', lol

 

the things is I turn those log base to 10

 

so, using my calculators.... = 6 the answer???

[imatfaal]

ok I am giving you too much help - but here goes, from here on though it hints only ;-)

 

your answer is correct but clumsy ( and that line starting with "so ..." is just plain bad!!!) - but this is how I would do it

 

[math] log_2(9)*log_3(4)*log_4(8) [/math]

 

[math] \frac{log9}{log2}*\frac{log4}{log3}*\frac{log8}{log4}[/math] change to a single base

 

[math] \frac{log9}{log3}*\frac{log4}{log4}*\frac{log8}{log2} [/math] rearrange

 

[math] 2 * 1 * \frac{log8}{log2} [/math] simplify the obvious division of logs to numbers

 

[math] \frac{2*log(8)}{log(2)} = \frac{2*log(2^3)}{log(2)} = \frac{2*3*log(2)}{log(2)} = \frac{6 log(2)}{log(2)} = 6 \frac{log2}{log2} = 6 [/math]

 

No need for a calculator. I haven't specified which base - because with these operations it doesn't matter, at no point did I need to evaluate a log

 

 

If I was helpful, let me know by clicking the [+] sign ->

Edited May 18, 2011 by imatfaal

[Vastor]

ok I am giving you too much help - but here goes, from here on though it hints only ;-)

 

your answer is correct but clumsy ( and that line starting with "so ..." is just plain bad!!!) - but this is how I would do it

 

[math] log_2(9)*log_3(4)*log_4(8) [/math]

 

[math] \frac{log9}{log2}*\frac{log4}{log3}*\frac{log8}{log4}[/math] change to a single base

 

[math] \frac{log9}{log3}*\frac{log4}{log4}*\frac{log8}{log2} [/math] rearrange

 

[math] 2 * 1 * \frac{log8}{log2} [/math] simplify the obvious division of logs to numbers

 

[math] \frac{2*log(8)}{log(2)} = \frac{2*log(2^3)}{log(2)} = \frac{2*3*log(2)}{log(2)} = \frac{6 log(2)}{log(2)} = 6 \frac{log2}{log2} = 6 [/math]

 

No need for a calculator. I haven't specified which base - because with these operations it doesn't matter, at no point did I need to evaluate a log

 

 

If I was helpful, let me know by clicking the [+] sign ->

 

ermm.. so u can do 'rearrange' part???

 

 

the point is, i never know that, that's why i use calculator, thnx for the calculation and yes, it's helpful

[kavlas]

i think i should move to the 'real question', because [math] log_3(9) [/math] is just my creation. So it's said

 

Evaluate [math] log_2 9 * log_3 4 * log_4 8 [/math] 2marks

 

 

if I'm not wrong, converting log base would be using this formula

 

[math] log_a(b) = \frac {log_c b}{log_c a} [/math]

 

so [math] log_2(9) = \frac {log_10(9)}{log_10(2)} [/math] * [math] log_3(4) = \frac {log_10(4)}{log_10(3)} [/math] * [math] log_4(8) = \frac {log_10(8)}{log_10(4)} [/math]

 

don't know how to 'zoom out' the '0', lol

 

the things is I turn those log base to 10

 

so, using my calculators.... = 6 the answer???

 

[math] log_2 9 * log_3 4 * log_4 8 [/math]= [math]log_2 3^2*log_3 2^2*log_4 2^3=2log_2 3*2log_3 2*3\frac{1}{log_2 4}[/math]= [math] 12*\frac{1}{2log_2 2} = 6[/math] by using the formula :

 

................................[math]log_a b*log_b a = 1[/math]

[DJBruce]

ermm.. so u can do 'rearrange' part???

 

 

the point is, i never know that, that's why i use calculator, thnx for the calculation and yes, it's helpful

 

Yes, you can do the rearranging part since [math]log(a)[\math] just represents a real number you still have the commutativity of multiplication.

[Vastor]

I think it's not relevant to start new topic about this...

 

ok let's start.. this is the law of log that i learn at school :-

 

 

[math] log_a mn = log_a m + log_a n [/math]

 

[math]log_a\frac{m}{n} = log_a m - log_a n[/math]

 

[math]log_a m^n = n log_a m [/math]

 

[math]log_a a = 1[/math]

 

[math]log_a 1 = 0[/math]

 

[math]log_a b = \frac{log_c b}{log_c a}[/math]

 

[math]log_a b = \frac{1}{log_b a}[/math]

 

 

there any more laws? how about law of multiply and dividing between log???

[Shadow]

Nope. There's a nice identity I found when learning log's, [math]\sqrt[\log_y{x}]{x} = y[/math], although it's pretty much useless as far as I know. The only thing you can do with multiplication and/or division of logarithms is use the rules you already know; [math]\log{a} \cdot \log{b} = \log{a^{\log{b}}} = \log{b^{\log{a}}}[/math] and [math]\frac{\log{a}}{\log{b}} = \log{\sqrt[\log{b}]{a}} = \frac{1}{\log{\sqrt[\log{a}]{b}}} = \log_{\sqrt[\log{a}]{b}}{10}[/math].

[Vastor]

does

 

[math]log\sqrt{x^3} = \frac{3}{2} log x[/math] ??

 

just need confirmation of my understanding

Edited May 31, 2011 by Vastor

[Shadow]

Yes.