Question

O valor de log 64/81 em função de a e b , sendo log 2 = a e log 3 = b, é:

6a + 4b

4b - 6a

6a -4b

4a - 6b

24ab

Answer 1

$\large\boxed{\begin{array}{l}\sf \ell og\bigg(\dfrac{64}{81}\bigg)=\ell og64-\ell og81\\\sf \ell og\bigg(\dfrac{64}{81}\bigg)=\ell og2^6-\ell og3^4\\\sf \ell og\bigg(\dfrac{64}{81}\bigg)=6\ell og2-4\ell og3\\\sf \ell og\bigg(\dfrac{64}{81}\bigg)=6a-4b\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~c}}}}\end{array}}$