Question

If log2= a and log3= b evaluate log(0.375)?

Answer 1

`log0.375=b-3a`

Explanation:

Note that `log5=log(10/2)=log10-log2=(1-a)`

We will now work on `log0.375`.

`log0.375`

= `log(375/1000)`

= `log375-log1000`

= `log(3xx5xx5xx5)-3`

= `log3+3log5-3`

= `b+3(1-a)-3`

= `b+3-3a-3`

= `b-3a`

Alternatively , one can also write

`log0.375=log(3/8)=log3-log8=log3-log2^3`

= `log3-3log2=b-3a`