[Math] Simplification of different base logarithms

Question

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

Answer 1

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 $$ \begin{align} \log_26-\log_49&=\log_2(2\cdot3)-\log_4(3^2) \\ &= \log_22+\log_23-2\log_43 &\text{using the first and third identities}\\ &=1+\log_23-2\log_43 &\text{using the sixth identity}\\ &=1+\log_23-2\log_42\log_23 &\text{using the fifth}\\ &=1+\log_23-2(\log_23)/\log_24 &\text{using the fourth}\\ &=1+\log_23-2(\log_23)/2 &\text{using the third}\\ &=1+\log_23-\log_23 &\text{using a bit of algebra}\\ &=1 \end{align} $$