Question

Without using tables , evaluate
2log5+log3+3log2-1/2log36-2log10

Answer 1

Answer:

Step-by-step explanation:

2 log 5 + log 3 + 3 log 2 - log(36)^(1 / 2) - 2 log 5×2 =

2 log 5 + log 3 + 3 log 2 - log 6 - 2(log 5 + log 2) =

2 log 5 + log 3 + 3 log 2 - log(2×3) - 2 log 5 - 2log 2 =

2log 5 + log 3 + 3log 2 - log 2 - log 3 - 2 log 5 - 2 log 2 = 0.

the answer is 0.