Hi,
2log(x) + 5 = 1 |-5
2log(x) = -4
log(x) = -2 |10 anwenden (also Umkehrfunktion des Logarithmus
10^{log x} = 10^{-2}
x = 10^{-2} = 0,01
log(x) + 2log(10x) = 8 |a*log(x) = log(x^a)
log(x) + log((10x)^2) = 8 |log(a) + log(b) = log(ab), sowie (10x)^2 = 100x^2
log(100x^3) = 8 |10 anwenden
100x^3 = 10^8 |:100
x^3 = 10^6 |Dritte Wurzel
x = 10^2 = 100
log(x) + log(52-x) = 2 |log(a) + log(b) = log(ab)
log(x*(52-x)) = 2 |10 anwenden
52x-x^2 = 100 |pq-Formel
x_(1) = 2 und x_(2) = 50
Probe bestätigt beide Ergebnisse
Grüße