Question

Given log630 ≈ 1.898 and log62 ≈ 0.387, log615 =

Answer 1

`\log_630\approx1.898\\\\\log_62\approx0.387\\\\\log_615=\log_6(30)/(2)=\log_630-\log_62\approx1.898-0.387=1.511\\\\Used:\\\\\log_ab-\log_ac=\log_a(b)/(c)`

Answer 2

The value of
`log_6{15}` is 1.511.

Explanation:

It is given that
`log_6{30}\approx 1.898` and
`log_6{2}\approx 0.387`.

We have to find the value of
`log_6{15}`.

It can be written as

`log_6{15}=log_6(30)/(2)`

Using property of lo, we get

`log_6{15}=log_630-log_62`
`log_m(a)/(b)=log_ma-log_mb`

Substitute
`log_6{30}\approx 1.898` and
`log_6{2}\approx 0.387` in the above equation.

`log_6{15}=1.898-0.387`

`log_6{15}=1.511`

Therefore the value of
`log_6{15}` is 1.511.