^-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
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
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
