Question

How do you solve `Log(x-9) = 3-Log(100x)`?

Answer 1

`color(blue)(x=10)`

Explanation:

`log(x-9)=3-log(100x)`

By the laws of logarithms:

`log(ab)=log(a)+log(b)color(white)(888)[1]`

`log(100x)=log(100)+log(x)`

Assuming these are base 10 logarithms:

`log(100)+log(x)=2+log(x)`

We now have:

`log(x-9)=3-2-log(x)`

`log(x-9)=1-log(x)`

Using `[1]`

`log(x-9)+log(x)=1`

`log(x(x-9))=1`

`log(x^2-9x)=1`

`10^(log(x^2-9x))=10^(1)`

`x^2-9x=10`

`x^2-9x-10=0`

Factor:

`(x+1)(x-10)=0=>x=-1 and x=10`

Checking solutions.

`x=-1`

`log((-1)-9)=3-log(100(-1))`

`log(-10)=3-log(-100)`

Logarithms are only defined for real numbers if for:

`log(x)`

`x>0`

Therefore `-1` is not a solution.

For `x=10`

`log(10-9)=3-log(100(10))`

`log(1)=3-log(1000)`

`0=3-3`

`0=0`

So `x=10` is the only solution.