Q.
Let B,C,P and L be positive real numbers such that log(B⋅L)+log(B⋅P)=2;log(P⋅L)+log(P⋅C)=3;log(C⋅B)+log(C⋅L)=4 The value of the product (BCPL) equals (base of the log is 10 )
108
102
Continuity and Differentiability
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Solution:
Given 2logB+logL+logP=2 ....(1) 2logP+logL+logC=3…..(2) and 2logC+logB+logL=4.....(3) add log(B3C3P3L3)=9 ⇒log(BCPL)=3⇒BCPL=1000