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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