Express logarithm in another variable

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The discussion revolves around expressing the logarithm log_6 3 in terms of variables m and n. Participants confirm that the expression can be manipulated using the relationships log_6 3 = 1 - log_6 2 and log_6(2·3) = 1. By creating two equations involving log_6 2 and log_6 3, the goal is to isolate the logarithm in terms of m and n. The approach involves adding log_6 3 to both sides to facilitate the solution. The conversation emphasizes the importance of completing the equations to derive the final expression.
songoku
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Homework Statement
Given that ##\log_{6} 30=m## and ##\log_{6} 20=n##, express ##\log_{6} 3## in ##m## or ##n##
Relevant Equations
logarithm property
The best I can get is:

$$\log_{6} 3=m-n+\log_{6} 2$$

Is it possible to get the final answer in terms of ##m## and ##n## only? If yes, I will try to do it again

Thanks
 
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You can make use of the below
\log_63=1-\log_62
\log_65+1=m
2\log_62+\log_65=n
 
Last edited:
what you did is correct. you need to complete it by using ##\log_6(2\cdot3)=1##, so that you will have two equations with two unknowns (unknowns can be considered to be the ##\log_6 2## and ##\log_6 3##.)
 
Try adding ##\log_6 3## to both sides.
 
Thank you very much anuttarasammyak, Delta2, vela
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.

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