Question

log 2 com base 3 dividido log 8 com base 7? gostaria de saber, por favor.

Answer 1

log3(2)
______
log7(8)


log3(2)
______
log7(2^3)  //  Log a (b^x) = x.loga(b)

log3(2)
______        =====>   ~ 0.590415
3log7(2)

Answer 2

Na base 7
$\dfrac{\log_3(2)}{\log_7(8)}=\dfrac{\log_3(2)}{\log_7(2^3)}=\dfrac{\log_3(2)}{3\log_7(2)}=\dfrac{\frac{\log_7(2)}{\log_7(3)}}{3\log_7(2)}=\dfrac{\log_7(2)}{\log_7(3)}\times\dfrac{1}{3\log_7(2)}=\dfrac{1}{\log_7(3)}\times\dfrac{1}{3}=\dfrac{1}{3\log_7(3)}=\boxed{\dfrac{1}{\log_7(27)}}$

Na base 3
$\dfrac{\log_3(2)}{\log_7(8)}=\dfrac{\log_3(2)}{\frac{\log_3(8)}{\log_3(7)}}=\log_3(2)\times{\dfrac{\log_3(7)}{\log_3(8)}}=\log_3(2)\times{\dfrac{\log_3(7)}{\log_3(2^3)}}=\log_3(2)\times{\dfrac{\log_3(7)}{3\log_3(2)}}=\boxed{\dfrac{\log_3(7)}{3}}}$