Vastor Posted May 17, 2011 Share Posted May 17, 2011 (edited) 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 Link to comment Share on other sites More sharing options...
DJBruce Posted May 17, 2011 Share Posted May 17, 2011 (edited) 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 Link to comment Share on other sites More sharing options...
Vastor Posted May 17, 2011 Author Share Posted May 17, 2011 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 ? Link to comment Share on other sites More sharing options...
imatfaal Posted May 17, 2011 Share Posted May 17, 2011 You sure DJB - looks fine to me. 2 Link to comment Share on other sites More sharing options...
Vastor Posted May 17, 2011 Author Share Posted May 17, 2011 and there are a law of logs [math] log_a a = 1[/math] Link to comment Share on other sites More sharing options...
DJBruce Posted May 17, 2011 Share Posted May 17, 2011 My bad, wasn't quite thinking there. Link to comment Share on other sites More sharing options...
imatfaal Posted May 17, 2011 Share Posted May 17, 2011 logx(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) 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 17, 2011 Author Share Posted May 17, 2011 logx(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 Link to comment Share on other sites More sharing options...
Vastor Posted May 18, 2011 Author Share Posted May 18, 2011 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 Link to comment Share on other sites More sharing options...
imatfaal Posted May 18, 2011 Share Posted May 18, 2011 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 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 18, 2011 Author Share Posted May 18, 2011 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??? Link to comment Share on other sites More sharing options...
imatfaal Posted May 18, 2011 Share Posted May 18, 2011 (edited) 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 2 Link to comment Share on other sites More sharing options...
Vastor Posted May 19, 2011 Author Share Posted May 19, 2011 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 Link to comment Share on other sites More sharing options...
kavlas Posted May 19, 2011 Share Posted May 19, 2011 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] Link to comment Share on other sites More sharing options...
DJBruce Posted May 19, 2011 Share Posted May 19, 2011 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. Link to comment Share on other sites More sharing options...
Vastor Posted May 29, 2011 Author Share Posted May 29, 2011 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??? Link to comment Share on other sites More sharing options...
Shadow Posted May 30, 2011 Share Posted May 30, 2011 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]. 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 31, 2011 Author Share Posted May 31, 2011 (edited) does [math]log\sqrt{x^3} = \frac{3}{2} log x[/math] ?? just need confirmation of my understanding Edited May 31, 2011 by Vastor Link to comment Share on other sites More sharing options...
Shadow Posted May 31, 2011 Share Posted May 31, 2011 Yes. 1 Link to comment Share on other sites More sharing options...
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