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April 25th, 2011, 07:38 PM  #1 
Newbie Joined: Dec 2010 Posts: 12 Thanks: 0  Log question (need help quick)
Hi everyone :P i got stuck while i'm trying to solve this question: Q8 Find a relation between x and y that does not involve logarithms: b) 2log10(y)  3log10(x) = 2 <by the way, 10 is the base> it'll b a great help if anyone can show me the process of solving this question thank you 
April 25th, 2011, 07:41 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs  Re: Log question (need help quick)
We have: 
April 25th, 2011, 08:14 PM  #3 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Log question (need help quick)
It takes my browser longer to download the webpage than it did for you to reply!

April 25th, 2011, 08:34 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs  Re: Log question (need help quick) I gave no explanations, as it seemed pretty evident what was being done, that's how I was so fast. 
April 26th, 2011, 03:08 PM  #5 
Newbie Joined: Dec 2010 Posts: 12 Thanks: 0  Re: Log question (need help quick)
thank you for your help. as you can see, i'm pretty new to logs could you help me solve this one as well? log10(x) = log10(4) / log10(2) the problem is i don't really get the concept of logs. i've only learnt the 5 basic laws of log and the division of logs is just out of my understanding. could you explain it for me? 
April 26th, 2011, 03:36 PM  #6 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Log question (need help quick)
Surely one of the rules that you know will justify the equality... log_10(4) = 2*log_10(2) Then the right hand side is just "2". Exponentiate. etc. 

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