1.

If log 2=0.301 and log of 0.4771 find the value of log(2430÷64)

Answer»

ANSWER:

Given LOG2=0.3010 and LOG3=0.4771

log12=log(4×3)

=log4+log3

=log2

2

+log3

=2log2+log3

Substituting the VALUES

=2×0.3010+0.4771

=0.602+0.4771

=1.0791



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