Question

How do you solve `log_8(11-6x)=log_8(1-x)`?

Answer 1

`{O/}`

Explanation:

`log_8(11 - 6x) - log_8(1 - x) = 0`

You need use the apply the rule `log_an - log_am = log_a(n/m)`.

`log_8((11 - 6x)/(1 - x)) = 0`

`(11- 6x)/(1 - x) = 8^0`

`11 - 6x = 1`

`-6x = -10`

`x =5/3`

However, checking in the original equation, you will realize that this solution is extraneous, because if in `log_b(n)`, `n <=0`, the equation will be undefined.

Hence, the solution set is `{O/}`.

Hopefully this helps!