Question

Given Log3=0.4772, log7=0.8451, log5=0.6990, log2=0.3010, how do you evaluate Log21?

Answer 1

`1.3223`

Explanation:

`\Log(21)`

`=\log(3\times 7)`

`=\log3+\log7`

`=0.4772+0.8451`

`=1.3223`