How do you approximate `log_5 (10/9)` given `log_5 2=0.4307` and `log_5 3=0.6826`?

The Logarithmic Division Rule states that:

`log_b(x/y)=log_b(x)-log_b(y)`

The Logarithmic Multiplication Rule states that:

`log_b(x*y)=log_b(x)+lob_b(y)`

We can apply the logarithmic division rule, so:

`log_5(10/9)`

becomes:

`log_5(10)-log_5(9)`

We can simplify the numbers in the brackets into the products of prime numbers:

`->log_5(2*5)-log_5(3*3)`

Now, we can apply the logarithmic multiplication rule, so:

`->(log_5(2)+log_5(5))-(log_5(3)+log_5(3))`

Now, we can substitute in the values given to us:

`->(0.4307+1)-(0.6826+0.6826)`

`(log_b(b)=1)`, always

`=(1.4307)-(1.3652)`

`=0.0655`

We apply

1. The Logarithmic Division Rule which states that

   `log(x/y)=logx-logy`

2. The Logarithmic Multiplication Rule

   `log(x*y)=logx+logy`

3. And the Logarithmic Power Rule

   `log(x^y)=ylogx`

Given number can be written as

`log_5(10/9)`
`=>log_5((2xx5)/3^2)`
`=>log_5(2xx5)-log_5 3^2` .......(Division Rule)
`=>log_5 2+log_5 5-log_5 3^2` .......(Multiplication Rule)
`=>log_5 2+log_5 5-2log_5 3` .......(Power Rule)

Inserting the given values and remembering that `log_b(b)` is always`=1` we get

`log_5(10/9)=0.4307+1-2xx0.6826`
`=>log_5(10/9)=0.0655`