logarytm proszę o pomoc Monika: Kolejny logarytm... 10* 100^{1₂ log9−log2}

Basia: 100 = 10^{2 1}₂log9 − log2 = log9^1/2 − log2 = log3 − log2 = log³₂ i masz 10*(10²)^log3₂ = 10*10^2log3₂ = 10*10^log(3₂)2 = 10*10^log9₄ = 10*⁹₄ = ^5*9₂ = ⁴⁵₂ = 22,5 bo a^log_ab = b, a log oznacza log₁₀

Vogl: 10*100^{0.5*log 9 − log 2}=10*100^{log 90.5 − log 2}= =10*100^{log 3 − log 2} = 10*100^{log (3/2)}=100^0.5*100^{log (3/2)}= =100^{0.5+log (3/2)} = 100^{log √10 + log (3/2)} = 100^{log ((3√10)/2)} = = 10^{2 log ((3√10)/2)} = 10^{log 22.5} = 22.5

ostr: Log₃ 54 −log₃ 2+100 log³

daras: a ten ostatni skłądnik , to co ma za argument? https://matematykaszkolna.pl/strona/218.html