Question

How do you solve `log_3 x - log_3 (x-2) = log_3 48`?

Answer 1

I found: `x=2.0425`

Explanation:

We can use the property of logs dealing with subtraction as:
`logx-logy=log(x/y)`
and get:
`log_3(x/(x-2))=log_3(48)`
the logs are equal when their arguments are equal so we get that the argments of the logs must be equal too:
`x/(x-2)=48`
rearranging:
`x=48(x-2)`
`x=48x-96`
`48x-x=96`
`47x=96`
`x=96/47=2.0425`

We can understan the property:
`logx-logy=log(x/y)`
using an example:
first let us consider:
`log_3(81)-log_3(27)=4-3=1`
using the definition of log:
`log_3(9)=x ->3^x=9` so that `x=2`

Now let us try with the division:
`log_3(81/27)=log_3(3)=1`