Math

Evaluate
6 log2 ∛64 +10log3 (243)^1/5)

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  1. ∛64
    = 64^(1/3)
    = 4 = 2^2

    243^1/5 = 3

    so 6 log2 ∛64 +10log3 (243)^1/5)
    = 6log2 (2^2) + 10 log3 (3)
    = 6(2 log2 2) + 10 log3 3
    = 12 (1) + 10(1)
    = 22

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  2. 6log_2(64)^1/3+10log_3(243)^1/5
    =2log_2 (64)+10log_3 (243)^1/5
    =2log_2(2)^6+2log_3 (243)
    =2log_2(2)^6+2log_3(3)^5
    =2log_2(2)^6+10log_3(3)
    =12log_2(2)+10log_3(3)
    Being that log_a(a)=1
    Therefore 12(1)×10(1)
    =120

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