Question

How do you evaluate `log 1500`?

Answer 1

`2log2+log3+3log5`

Explanation:

Well we have

`log1500=log(15*100)=log15+log100`

`=log(3*5)+log10^2`

`=log3+log5+2log10`

`=2log10+log3+log5`

`=2log(2*5)+log3+log5`

`=2log2+log3+3log5`

Answer 2

If `log` here is common log (base 10), then `log 1500 = 2+log15`

Explanation:

If `log 1500` means `log_10 1500`, then we have:

`log 1500 = log(100*15)`

`= log100 + log 15`

`= log10^2 + log 15`

`= 2 + log 15`

If you are using tables of logarithms, you'll need scientific notation:

`log 1500 = log(1.5 * 10^3)`

`= 3 + log(1.5)`

`= 3 + 0.1761` (from table)

`= 3.1761`