Odpowiedzi

2010-03-31T23:51:59+02:00
^-ten znak znaczy ze jest w potendze
a=log8/log81
a=3/4(log2/log3)
a=3/4(logprzy podstawie 3 z 2)
b=1/log64
1=log prz podstawie 10 z 10
b=1/3(log prz podstawie 10 z 10/log4)
b=1/3(logprzy podstawie 4 z 10)
27⁴a + 16³b
27=3^3
16=4^2
3^9(logprzy podstawie 3 z 2) + 4^2(logprzy podstawie 4 z 10)=
2^9+10^2=4 512+100=612
tyle:D
2 4 2
  • Roma
  • Community Manager
2010-04-01T00:11:49+02:00
I sposób
a = log 8/log 81 = log 2³/log 3⁴ = 3*log 2/4*log 3 = ¾ * log 2/log3 = ¾ * log₃ 2
b =1/log 64 = 1/log 2⁶ = 1/6 * log 2 = ⅙ * log₂ 10

27^4a + 16^3b = (3³)^4 * ¾ * log₃ 2 + (2⁴)^3 * ⅙ * log₂ 10 = 3^3 * 4 * ¾ * log₃ 2 + 2^4 * 3 * ⅙ * log₂ 10 = 3^9 * log₃ 2 + 2^2 * log₂ 10 = (3^log₃ 2)⁹ + (2^log₂ 10)² = 2⁹ + 10² = 512 + 100 = 612

II sposób
a = log 8/log 81 = log₈₁ 8
b = 1/log 64 = log₆₄ 10

(3³)⁴ = (3⁴)³
27⁴ = 81³
27^4a = 81^3a
81^3a = 81^3 * log₈₁ 8 = 81^log₈₁ 8³ = 8³ = 512
27^4a = 512

(4²)³ = (4³)²
16³ = 64²
16^3b = 64^2b
64^2b = 64^2 * log₆₄ 10 = 64^log₆₄ 10² = 10² = 100
16^3b = 100

27^4a + 16^3b = 512 + 100 = 612
7 4 7
2010-04-01T00:35:57+02:00
A= log8/log81 = log2³/log3⁴ = 3log2/(4log3) = ¾log2/log3 = ¾log₃2
b=1/log64 = 1/log2⁶ = 1/(6log2)

^ = potęgowanie
27^(4a) = 3³ ^ (4*¾log₃2)= 3^(9log₃2) = 3^(9log₃2)
16^(3b) = 2⁴ ^[3/(6log2)] = 2^(2/log2) = 2^(log100/log2) = 2^log₂100
Podstawmy:
x = 3^(9log₃2)
log₃x = 9log₃2
log₃x = log₃2⁹
x = 2⁹ = 512

y = 2^log₂100
log₂y = log₂100
y = 100

Ostatecznie:
27^(4a) + 16^(3b) = 512 + 100 = 612
1 5 1