Simplification of different base logarithms I'm in doubt on simplifying the expression:
$\log_2 6 - \log_4 9$
Working on it I've got: $\log_2 6 - \dfrac{\log_2 9}{2}$
There's anyway to simplify it more ? I'm learning logarithms now so I'm not aware of all properties and tricks.
Thanks in advance
 A: You can do a bit more simplification. The important properties of the log function are, for any base $a>0$,


*

*$\log_a(bc)=\log_ab+\log_ac$, so, for example, $\log_26=\log_22+\log_23$

*$\log_a(b/c)=\log_ab-\log_ac$

*$\log_ab^n=n\log_ab$

*$\log_ab=1/\log_ba$

*$(\log_ab)(\log_bc)=\log_ac$

*$\log_aa=1$


So, for example, we can simplify $\log_26-\log_49$ as
$$
\begin{align}
\log_26-\log_49&=\log_2(2\cdot3)-\log_4(3^2) \\
 &= \log_22+\log_23-2\log_43 &\text{using the first and third identities}\\
 &=1+\log_23-2\log_43 &\text{using the sixth identity}\\
 &=1+\log_23-2\log_42\log_23 &\text{using the fifth}\\
 &=1+\log_23-2(\log_23)/\log_24 &\text{using the fourth}\\
 &=1+\log_23-2(\log_23)/2 &\text{using the third}\\
 &=1+\log_23-\log_23 &\text{using a bit of algebra}\\
 &=1
\end{align}
$$
A: You're correct so far. Bring out the factor of $\frac{1}{2}$ to get $\frac{1}{2}(2\log_{2}(6)-\log_{2}(9))$. Since $a\log(b)=\log(b^{a})$ $\log(a)-\log(b)=\log(\frac{a}{b})$, you get $2\log_{2}(6)=\log_{2}(6^{2})$, and $$\frac{1}{2}(\log_{2}(36)-\log_{2}(9))=\frac{1}{2}\left(\log_{2}\left(\frac{36}{9}\right)\right)=\log_{2}(4)/2=2/2=1 $$
A: $log_26-\frac{log_29}{2}=log_22+log_23-1/2×（2×log_23）=1$
