log base 16 of 49 - log base 4 of 7. Explain

Question

hero · Answer

Use the change of base formula to simplify this

anonymous · Answer

$$(\log49/\log16)-(\log7/\log4)$$
Then what?

hero · Answer

You have to use change of base formula appropriately to simplify this. 

Hint:
$$\log_{7} 7^2 = 2$$

anonymous · Answer

But there are no bases that work with 49 and 16 or 7 and 4. You're left with: $$\log_{4}49/2 - \log_{2}7/4 $$

hero · Answer

$$\frac{\log_4{49}}{\log_4{4^2}} = \frac{1}{2}\log_4{49}$$

Now you have:

$\dfrac{1}{2}\log_4{7^2} - \log_4{7} = \dfrac{2}{2}\log_4{7} - \log_4{7} $
                             
                             $ = 0$

hero · Answer

Do you get it now?

anonymous · Answer

Sort of... The general sense. Thanks.

anonymous · Answer

All set!