помогите решить ети два примера вверху

там во втором примери 5^log1/3 9-log1/9 9

loga xⁿ = n loga x

logaⁿ x = 1/n loga x

1) log1/3 27 - log1/9 27

  log1/3 27 = log1/3 3³ = 3log1/3 3 = 3 * (-1) = -3

  log1/9 27 = log3⁻² 3³ = 3/(-2) * log3 3 = -1.5

  -3 - (-1.5) = -1.5

2) 5^log1/3 9-log1/9 9

   log1/3 9 = log3⁻¹ 3² = 2/(-1) * log3 3 = -2

   log1/9 9 = log9⁻¹ 9 = 1/(-1) * log9 9 = -1

  5^(-2-(-1) = 5^(-1) = 1/5