Question

How do you solve `log x- log(x-8)=1`?

Answer 1

`x=80/9=8 8/9`

Explanation:

The given:`log x- log (x-8) = 1`

This means the common logarithm of base 10:

`log_10 x-log_10(x-8) = 1`

Also `log_10 10=1`

therefore

`log_10 x-log_10(x-8) = log_10 10`

`log_10 (x/(x-8))=log_10 10`

Take Antilogarithm of both sides of the equation

`Antilog(log_10 (x/(x-8)))=Antilog(log_10 10)`

means

`10^(log_10 (x/(x-8)))=10^(log_10 10)`

means

`x/(x-8)=10`

solve for `x` now

`x=10(x-8)`

`x=10x-80`

`x-10x=-80`

`-9x=-80`

`(-9x)/-9=(-80)/-9`

`x=80/9=8 8/9`

Have a nice day !!! from the Philippines..