To evaluate the equation `log_6(3x-10)=log_6(14-5x)` , we apply logarithm property: `a^(log_a(x))=x` .

Raised both sides by base of `6` .

`6^(log_6(3x-10))=6^(log_6(14-5x))`

`3x-10=14-5x`

Add `10` on both sides.

`3x-10+10=14-5x+10`

`3x=24-5x`

Add `5x ` on both sides.

`3x+5x=24-5x+5x`

`8x=24`

Divide both sides by `8` .

`(8x)/8=24/8`

`x=3`

Checking: Plug-in `x=3` on `log_6(3x-10)=log_6(14-5x)` .

`log_6(3*3-10)=?log_6(14-5*3)`

`log_6(9-10)=?log_6(14-15)`

`log_3(-1)=?log_3(-1) FALSE`

A logarithm `log_b(x)` is undefined for `xlt=0` .

Thus, the `x=3` is an extraneous solution of the given equation
`log_6(3x-10)=log_6(14-5x)` . There is **no real
solution**.

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