Changing the base of a logarithm

[Jouster]

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

 

[pasmith]

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.

 

[Mark44]

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.

 

[Jouster]

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

 

[symbolipoint]

The Change Of Base Formula, as learned in Intermediate Algebra course?

 

[symbolipoint]

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.

 

[renormalize]

But I don't understand how to use this information & solve the problem

Some useful hints: $\log_{2}4=\log_{2}2^{2}$, $\log_{2}8=\log_{2}2^{3}$, $\log_{2}6=\log_{2}\left(2\times3\right)$, $\log_{2}12=\log_{2}\left(2^{2}\times3\right)$.

 

[pasmith]

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

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