# 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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