Changing the base of a logarithm

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To express \log_4(12) in terms of p, start by rewriting \log_8(6) as p, which leads to the equation 6 = 8^p. Converting to base 2, where 8 = 2^3 and 4 = 2^2, allows for the expression of both logarithms in terms of \log_2. The Change of Base Formula can be applied, leading to log_4(12) being expressed as a combination of \log_4(2) and \log_4(6). Ultimately, the relationship can be simplified to log_4(12) = (1/p)(log_4(2) + log_4(6)), although careful verification of each step is recommended.
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Homework Statement
I'm unsure of how to solve this problem, I tried changing the bases but they don't seem to have any similarities
Relevant Equations
Expressing values in terms of another
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Jouster said:
Homework Statement:: I'm unsure of how to solve this problem, I tried changing the bases but they don't seem to have any similarities

Given \log_8(6) = p, express \log_4(12) in terms of p.

Given that 8 = 2^3 and 4 = 2^2, expressing everything in terms of \log_2 seems like a good idea.
 
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The equation ##\log_8(6) = p## can be rewritten as ##6 = 8^p##. It's not too difficult to get from there to an equation involving ##\log_4(12)## that involves p.
 
pasmith said:
Given that 8 = 2^3 and 4 = 2^2, expressing everything in terms of \log_2 seems like a good idea.
I did express everything in terms of log2, I got
log8(6) = log26/log28 = p
log4(12) = log212/log24
But I don't understand how to use this information & solve the problem :biggrin:
 
The Change Of Base Formula, as learned in Intermediate Algebra course?
 
I tried but without using Change Of Base formula.
One step I found p*log_4(8)=log_4(6)
(Excuse the lack of great typesetting.)
and notice that 12=1.5*8, and 12=2*6; and continued on that way.

Lastly I found log_4(12)=(1/p)(log_4(2)+log_4(6).
Not sure if that is what was wanted. Also not absolutely sure it is correct - not feel like checking carefully.
 
Jouster said:
But I don't understand how to use this information & solve the problem :biggrin:
Some useful hints: ##\log_{2}4=\log_{2}2^{2}##, ##\log_{2}8=\log_{2}2^{3}##, ##\log_{2}6=\log_{2}\left(2\times3\right)##, ##\log_{2}12=\log_{2}\left(2^{2}\times3\right)##.
 
Jouster said:
I did express everything in terms of log2, I got
log8(6) = log26/log28 = p
log4(12) = log212/log24
But I don't understand how to use this information & solve the problem :biggrin:

But you didn't notice that \log_2 8 =\log_2(2^3) =3 and \log_2(4) = 2, or that \log_2(12) = \log_2(2 \times 6) = 1 + \log_2(6)?
 
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