`log_6(3x)+log_6(x-1)=3` Solve the equation. Check for extraneous solutions.

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To evaluate the given equation `log_6(3x)+log_6(x-1)=3` , we may apply the logarithm property: `log_b(x)+log_b(y)=log_b(x*y)` .




To get rid of the "log" function, we may apply the logarithm property: `b^(log_b(x))=x` .

Raise both sides by base of `6` .



Subtract `216` from both sides of the equation to simplify in standard form: `ax^2+bx+c= 0` .



Apply factoring on the trinomial.

`3*(x + 8)*(x - 9)=0`

Apply zero-factor property to solve for `x` by equating each factor in terms of `x` to `0` .







Checking: Plug-in each `x` on  `log_6(3x)+log_6(x-1)=3` .

Let `x=-8` on `log_6(3x)+log_6(x-1)=3` .



undefined +undefined =?3     FALSE

Note that `log_b(x)` is undefined on `xlt=0` .

Let `x=9` on `log_6(3x)+log_6(x-1)=3` .








`3=3 `    TRUE


Thus, the `x=-8` is an extraneous solution.

The `x=9` is the only real solution of the equation `log_6(3x)+log_6(x-1)=3` .

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