Odpowiedz Jola: Oblicz (abc) ^1/3, jeżeli log₂ a=5, b=log0,01, c=log_0,05 20

Saizou : a=2⁵=32 b=log0,01=log10⁻²=−2 c=log_5/10020=log_1/2020=log_20⁻¹20=−1 (abc)^1/3=(32*(−2)*(−1))^1/3=(2⁶)^1/3=2²=4

Basia: log₂a=5 ⇔ a=2⁵

	1
log0,01 = b ⇔ 10b = 0,01 =		= 10−2 ⇔ b= −2
	102

log_0,0520 = c ⇔ 0,05^c = 20 ⇔ (⁵₁₀₀)^c = 20 ⇔ (¹₂₀)^c = 20 ⇔ (20⁻¹)^c = 20¹ ⇔ 20^−c = 20¹ ⇔ −c=1 ⇔ c= −1 (abc)^1/3 = (2⁵*(−2)*(−1))^1/3 = (2⁶)^1/3 = 2² = 4