If log3=.4771, log5=.6990, then find the value of log150
Correct Answer: Description for Correct answer:
\( \Large log150=log \left(15 \times 10\right)=log15+log_{10}10 \)
=\( \Large log \left(5 \times 3\right)+1 \)
=\( \Large log5+log3+1 \)
=.6990+.4771+1
= 2.1761
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