`log_5(5x+9)=log_5(6x)` Solve the equation. Check for extraneous solutions.

To solve the equation `log_5(5x+9)=log_5(6x)` , we apply logarithm property: `a^(log_a(x))=x` .

Raise both sides by base of `5` .

`5^( log_5(5x+9))=5^(log_5(6x))`

`5x+9=6x`

Subtract `5x` from both sides of the equation

`5x+9-5x=6x-5x`

`9=x`

`x=9`

Checking: Plug-in `x=9` on `log_5(5x+9)=log_5(6x)` .

`log_5(5*9+9)=?log_5(6*9)`

`log_5(45+9)=?log_5(54)`

`log_5(54)=log_5(54)` TRUE

Thus, the `x=9` is the real exact solution of the equation . `log_5(5x+9)=log_5(6x)`. There is no extraneous solution.

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