Question

$3^{2- log_{3}10 }$ - $4^{1-log_{4}5 }$+$6^{log_{6}9-1 }$ =??

Answer 1

folosesti formula $a^{log _{a} x}$ =x.

$3^{log_{3} 9 - log _{3} 10 }$ - $4^{log_{4}4-log_{4}5 }$ + $6^{log _{6}9-log_{6}6 }$=
$3 ^{log_{3} \frac{9}{10} } -4^{log_{4} \frac{4}{5} } - 6^{log_{6} \frac{9}{6} } =$
$\frac{9}{10} - \frac{4}{5} + \frac{9}{6} =$
$\frac{48}{30} = \frac{24}{15}= \frac{8}{5}$