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