Question

Aratati ca numarul a=log3 8 *log4 9,e natural(*-inmultire).

Answer 1

$log_38*log_49$

$log_38=3log_32$
$log_49=2log_43$

$6log_32*log_43 = 6 \frac{ln 2}{ln 3}* \frac{ln 3}{ln 4}$
$6 \frac{ln2}{ln4} = \frac{6ln2}{2ln2} = \frac{6}{2} = 3$

Answer 2

am tinut cont ca
loga b * logb a=1