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)` .
`log_6(3x)+log_6(x-1)=3`
`log_6(3x*(x-1))=3`
`log_6(3x^2-3x)=3`
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` .
`6^(log_6(3x^2-3x))=6^3`
`3x^2-3x=216`
Subtract `216` from both sides of the equation to simplify in standard form: `ax^2+bx+c= 0` .
`3x^2-3x-216=216-216`
`3x^2-3x-216=0`
Apply factoring on the trinomial.
`3*(x + 8)*(x - 9)=0`
Apply zero-factor property to solve for...
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)` .
`log_6(3x)+log_6(x-1)=3`
`log_6(3x*(x-1))=3`
`log_6(3x^2-3x)=3`
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` .
`6^(log_6(3x^2-3x))=6^3`
`3x^2-3x=216`
Subtract `216` from both sides of the equation to simplify in standard form: `ax^2+bx+c= 0` .
`3x^2-3x-216=216-216`
`3x^2-3x-216=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` .
`x+8-8=0-8`
`x=-8`
`x-9=0`
`x-9+9=0+9`
`x=9`
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` .
`log_6(3*(-8))+log_6(-8-1)=?3`
`log_6(-24)+log_6(-9)=?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` .
`log_6(3*9)+log_6(9-1)=?3`
`log_6(27)+log_6(8)=?3`
`log_6(27*8)=?3`
`log_6(216)=?3`
`log_6(6^3)=?3`
`3log_6(6)=?3`
`3*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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